1. Use suitable identities to find the following products:
( i ) `\ \left( x + 4 \right)\left( x + 10 \right)\ `
( ii ) `\ \left( x + 8 \right)\left( x - 10 \right)\ `
( iii ) `\ \left( 3x + 4 \right)\left( 3x - 5 \right)\ `
( iv ) `\ \left( y^2 + \frac{3}{2} \right)\left( y^2 - \frac{3}{2} \right)\ `
( v ) `\ \left( 3 - 2x \right)\left( 3 + 2x \right)\ `
solution :
( i ) `\ \left( x + 4 \right)\left( x + 10 \right)\ `
Here we can use identity IV :
` \ ( x + a ) ( x + b ) = x^2 + ( a + b )x + ab \ `
` \ ( x + 4 ) ( x + 10) \ `
` \ = x^2 + ( 4 + 10 )x + 4×10 \ `
` \ = x^2 + 14x + 40 \ `
( ii ) `\ \left( x + 8 \right)\left( x - 10 \right)\ `
Here we can use identity IV :
` \ ( x + a ) ( x + b ) = x^2 + ( a + b )x + ab \ `
` \ ( x + 8 ) ( x - 10) \ `
` \ = x^2 + ( 8 + (-10) )x + 8×(-10) \ `
` \ = x^2 - 2x - 80 \ `
( iii ) `\ \left( 3x + 4 \right)\left( 3x - 5 \right)\ `
Here we can use identity IV :
` \ ( x + a ) ( x + b ) = x^2 + ( a + b )x + ab \ `
` \ ( 3x + 4 ) ( 3x - 5) \ `
` \ = (3x)^2 + ( 4 + (-5) )3x + 4×(-5) \ `
` \ = 9x^2 - 3x - 20 \ `
( iv ) `\ \left( y^2 + \frac{3}{2} \right)\left( y^2 - \frac{3}{2} \right)\ `
Here we can use identity III :
` \ ( x + a ) ( x - a ) = x^2 - a^2 \ `
` \ ( y^2 + \frac{3} {2} ) ( y^2 - \frac{3} {2}) \ `
` \ = (y^2)^2 - (\frac{3} {2})^2 \ `
` \ = y^4 - \frac{9} {4} \ `
( v ) `\ \left( 3 - 2x \right)\left( 3 + 2x \right)\ `
Here we can use identity III :
` \ ( x + a ) ( x - a ) = x^2 - a^2 \ `
` \ ( 3 - 2x ) ( 3 + 2x) \ `
` \ = 3^2 - (2x)^2 \ `
` \ = 9 - 4x^2 \ `
2. Evaluate the following products without multiplying directly:
( i ) `\ 103 \times 107\ `
( ii ) `\ 95 \times 96\ `
( iii ) `\ 104 \times 96\ `
solution :
( i ) `\ 103 \times 107\ `
` \ 103 × 107 = (100 + 3) (100 + 7) \ `
` \ = (100)^2 + (3 + 7)(100) + (3×7) \ ` using identity IV
` \ = 10000 + 1000 + 21 \ `
` \ = 11021 \ `
( ii ) `\ 95 \times 96\ `
` \ 95 × 96 = (100 + (-5)) (100 + (-4)) \ `
` \ = (100)^2 + ((-5) + (-4)) \ ` ` \ (100) + ((-5)×(-4)) \ `
using identity IV
` \ = 10000 + (-900) + 20 \ `
` \ = 10000 - 900 + 20\ `
` \ = 9120 \ `
( iii ) `\ 104 \times 96\ `
` \ 104 × 96 = (100 + 4) (100 - 4) \ `
` \ = (100)^2 - 4^2 \ ` using identity III
` \ = 10000 - 16 \ `
` \ = 9984 \ `
3. Factorise the following using appropriate identities:
( i ) `\ 9 x^2 + 6xy + y^2 \ `
( ii ) `\ 4 y^2 - 4y + 1 \ `
( iii ) `\ x^2 - \frac{ y^2 }{ 100 }\ `
solution :
( i ) `\ 9 x^2 + 6xy + y^2 \ `
Here you can see that
`\ 9 x^2 + 6xy + y^2 \ `
` \ 9x^2 = (3x)^2, \ ` ` \ y^2 = (y)^2, \ ` ` \ 6xy = 2 (3x) (y) \ `
Comparing the given expression with ` \ x^2 + 2xy + y^2 ,\ ` we observe that ` \ x = 3x \ ` and ` \ y = y \ `.
Using Identity I, we get
`\ 9 x^2 + 6xy + y^2 \ ` ` \ = ( 3x + y )^2 \ `
` \ = ( 3x + y ) ( 3x + y ) \ `
( ii ) `\ 4 y^2 - 4y + 1 \ `
Here you can see that
`\ 4 y^2 - 4y + 1 \ `
` \ 4y^2 = (2y)^2, \ ` ` \ 1 = (1)^2, \ ` ` \ 4y = 2 (2y) (1) \ `
Comparing the given expression with ` \ x^2 - 2xy + y^2 ,\ ` we observe that ` \ x = 2y \ ` and ` \ y = 1 \ `.
Using Identity II, we get
`\ 4 y^2 - 4y + 1 \ ` ` \ = ( 2y - 1 )^2 \ `
` \ = ( 2y - 1 ) ( 2y - 1 ) \ `
( iii ) `\ x^2 - \frac{ y^2 }{ 100 }\ `
Here you can see that
`\ x^2 - \frac{ y^2 }{ 100 }\ `
` \ x^2 = (x)^2, \ ` ` \ \frac {y^2} {100} = (\frac {y} {10})^2, \ `
Comparing the given expression with ` \ x^2 - y^2 ,\ ` we observe that ` \ x = x \ ` and ` \ y = \frac {y} {10} \ `.
Using Identity III, we get
`\ x^2 - \frac{ y^2 }{ 100 }\ ` ` \ = ( x + \frac {y} {10} )(x - \frac {y} {10}) \ `
4. Expand each of the following, using suitable identities:
( i ) `\ \left( x + 2y + 4z \right)^2 \ `
( ii ) `\ \left( 2x - y + z \right)^2 \ `
( iii ) `\ \left( -2x + 3y + 2z \right)^2 \ `
( iv ) `\ \left( 3a - 7b - c \right)^2 \ `
( v ) `\ \left( -2x + 5y - 3z \right)^2 \ `
( vi ) `\ \left[ \frac{1}{4}x - \frac{1}{2}b + 1 \right]\ `
solution :
( i ) `\ \left( x + 2y + 4z \right)^2 \ `
Comparing the given expression with ` \ ( x + 2y + 4z )^2 \ `, we find that
` \ x = x,\ ` `\ y = 2y \ ` and ` \ z = 4z \ `.
Therefore, using Identity V, we have
` \ ( x + 2y + 4z )^2 \ `
` \ = (x)^2 + (2y)^2 + (4z)^2 \ ` ` \ + 2 (x) (2y) + 2 (2y) (4z) + 2 (4z) (x) \ `
` \ = x^2 + 4y^2 + 16z^2 \ ` ` \ + 4xy + 16yz + 8xz \ `
( ii ) `\ \left( 2x - y + z \right)^2 \ `
Comparing the given expression with ` \ ( 2x - y + z )^2 \ `, we find that
` \ x = 2x,\ ` `\ y = -y \ ` and ` \ z = z \ `.
Therefore, using Identity V, we have
` \ ( 2x - y + z )^2 \ `
` \ = (2x)^2 + (-y)^2 + (z)^2 \ ` ` \ + 2 (2x) (-y) + 2 (-y) (z) + 2 (z) (2x) \ `
` \ = 4x^2 + y^2 + z^2 \ ` ` \ - 4xy - 2yz + 4xz \ `
( iii ) `\ \left( -2x + 3y + 2z \right)^2 \ `
Comparing the given expression with ` \ ( -2x + 3y + 2z )^2 \ `, we find that
` \ x = -2x,\ ` `\ y = 3y \ ` and ` \ z = 2z \ `.
Therefore, using Identity V, we have
` \ ( -2x + 3y + 2z )^2 \ `
` \ = (-2x)^2 + (3y)^2 + (2z)^2 \ ` ` \ + 2 (-2x) (3y) + 2 (3y) (2z) + 2 (2z) (-2x) \ `
` \ = 4x^2 + 9y^2 + 4z^2 \ ` ` \ - 12xy + 12yz - 8xz \ `
( iv ) `\ \left( 3a - 7b - c \right)^2 \ `
Comparing the given expression with ` \ ( 3a - 7b - c )^2 \ `, we find that
` \ x = 3a,\ ` `\ y = -7b \ ` and ` \ z = -c \ `.
Therefore, using Identity V, we have
` \ ( 3a - 7b - c )^2 \ `
` \ = (3a)^2 + (-7b)^2 + (-c)^2 \ ` ` \ + 2 (3a) (-7b) + 2 (-7b) (-c) + 2 (-c) (3a) \ `
` \ = 9a^2 + 49b^2 + c^2 \ ` ` \ - 42ab + 14bc - 6ac \ `
( v ) `\ \left( -2x + 5y - 3z \right)^2 \ `
Comparing the given expression with ` \ ( -2x + 5y - 3z )^2 \ `, we find that
` \ x = -2x,\ ` `\ y = 5y \ ` and ` \ z = -3z \ `.
Therefore, using Identity V, we have
` \ ( -2x + 5y - 3z )^2 \ `
` \ = (-2x)^2 + (5y)^2 + (-3z)^2 \ ` ` \ + 2 (-2x) (5y) + 2 (5y) (-3z) + 2 (-3z) (-2x) \ `
` \ = 4x^2 + 25y^2 + 9z^2 \ ` ` \ - 20xy - 30yz + 12xz \ `
( vi ) `\ \left[ \frac{1}{4}x - \frac{1}{2}b + 1 \right]\ `
Comparing the given expression with ` \ [ \frac {1} {4}a - \frac {1} {2}b + 1 ]^2 \ `, we find that
` \ x = \frac {1} {4}a,\ ` `\ y = - \frac {1} {2}b \ ` and ` \ z = 1 \ `.
Therefore, using Identity V, we have
` \ [ \frac {1} {4}a - \frac {1} {2}b + 1 ]^2 \ `
` \ = (\frac {1} {4}a)^2 + (-\frac {1} {2}b)^2 + (1)^2 \ ` ` \ + 2 (\frac {1} {4}a) (- \frac {1} {2}b) + 2 (- \frac {1} {2}b) (1) + 2 (1) (\frac {1} {4a}) \ `
` \ = \frac {1} {16}a^2 + \frac {1} {4}b^2 + 1 \ ` ` \ - \frac {1} {4}ab - b + \frac {1} {2}a \ `
5. Factorise:
( i ) `\ 4 x^2 + 9 y^2 + 16 z^2 \ ` ` \ + 12xy - 24yz - 16xz\ `
( ii ) `\ 2 x^2 + y^2 + 8 z^2 \ ` ` \ - 2\sqrt 2 xy - 4\sqrt 2 yz - 8xz\ `
solution :
( i ) `\ 4 x^2 + 9 y^2 + 16 z^2 \ ` ` \ + 12xy - 24yz - 16xz\ `
We have `\ 4 x^2 + 9 y^2 + 16 z^2 \ ` ` \ + 12xy - 24yz - 16xz\ `
` \ = (2x)^2 + (3y)^2 + (-4z)^2 \ ` ` \ + 2 (2x) (3y) + 2 (3y) (-4z) + 2 (-4z) (2x) \ `
⇒ ` \ [ 2x + 3y + (-4z) ]^2 \ ` ( Using Identity V )
⇒ ` \ ( 2x + 3y - 4z )^2 \ ` ` \ = ( 2x + 3y - 4z ) \ ` ` \ ( 2x + 3y - 4z ) \ `
( ii ) `\ 2 x^2 + y^2 + 8 z^2 \ ` ` \ - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz\ `
We have `\ 2 x^2 + y^2 + 8 z^2 \ ` ` \ - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz\ `
` \ = (- \sqrt 2 x)^2 + (y)^2 + (2 \sqrt 2 z)^2 \ ` ` \ + 2 (- \sqrt 2 x) (y) + 2 (y) (2 \sqrt 2 z) + 2 (2 \sqrt 2z) (- \sqrt 2 x) \ `
` \ [ (- \sqrt 2 x) + y + 2 \sqrt 2 z ]^2 \ ` ( Using Identity V )
` \ (- \sqrt 2 x + y = 2 \sqrt 2 z )^2 \ ` ` \ = (- \sqrt 2 x + y = 2 \sqrt 2 z ) \ ` ` \ (- \sqrt 2 x + y = 2 \sqrt 2 z ) \ `
6. Write the following cubes in expanded form:
( i ) `\ \left( 2x + 1 \right)^3 \ `
( ii ) `\ \left( 2a - 3b \right)^3 \ `
( iii ) `\ \left[ \frac{3}{2}x + 1 \right]^3 \ `
( iv ) `\ \left[ x + \frac{2}{3}y \right]^3 \ `
solution :
( i ) `\ \left( 2x + 1 \right)^3 \ `
Comparing the given expression with ` \ (x + y)^3 \ `, we find that
` \ x = 2x and y = 1. \ `
So, using Identity VI, we have:
` \ ( 2x + 1 )^3 \ ` ` \ = (2x)^3 + (1)^3 + 3 (2x) (1) ( 2x + 1 ) \ `
` \ 8x^3 + 1 + 6x ( 2x + 1 ) \ `
` \ 8x^3 + 1 + 12x^2 + 6x \ `
` \ 8x^3 + 12x^2 + 6x + 1 \ `
( ii ) `\ \left( 2a - 3b \right)^3 \ `
Comparing the given expression with ` \ (x + y)^3 \ `, we find that
` \ x = 2a and y = 3b. \ `
So, using Identity VII, we have:
`\ \left( 2a - 3b \right)^3 \ ` ` \ = (2a)^3 - (3b)^3 - 3 (2a) (3b) ( 2a - 3b ) \ `
` \ 8a^3 - 27b^3 - 18ab ( 2a - 3b ) \ `
` \ 8a^3 - 27b^3 - 36a^2b + 54ab^2 \ `
( iii ) `\ \left[ \frac{3}{2}x + 1 \right]^3 \ `
Comparing the given expression with ` \ (x + y)^3 \ `, we find that
` \ x = \frac {3} {2}x and y = 1. \ `
So, using Identity VI, we have:
`\ \left[ \frac{3}{2}x + 1 \right]^3 \ ` ` \ = (\frac {3} {2}x)^3 + (1)^3 + 3 (\frac {3} {2}x) (1) ( \frac {3} {2}x + 1 ) \ `
` \ \frac {27} {8}x^3 + 1 + \frac {9} {2}x ( \frac {3} {2}x + 1 ) \ `
` \ \frac {27} {8}x^3 + 1 + \frac {27} {4}x^2 + \frac {9} {2}x \ `
` \ \frac {27} {8}x^3 + \frac {27} {4}x^2 + \frac {9} {2}x + 1 \ `
( iv ) `\ \left[ x - \frac{2}{3}y \right]^3 \ `
Comparing the given expression with ` \ (x + y)^3 \ `, we find that
` \ x = x and y = \frac {2} {3}y . \ `
So, using Identity VII, we have:
`\ \left[ x - \frac{2}{3}y \right]^3 \ ` ` \ = (x)^3 - (\frac {2} {3}y)^3 - 3 (x) (\frac {2} {3}y) ( x - \frac {2} {3}y ) \ `
` \ x^3 - \frac {8} {27} y^3 - 2xy ( x - \frac {2} {3}y ) \ `
` \ x^3 - \frac {8} {27} y^3 - 2x^2y + \frac {4} {3}xy^2 \ `
7. Evaluate the following using suitable identities:
( i ) `\ \left( 99 \right)^3 \ `
( ii ) `\ \left( 102 \right)^3 \ `
( iii ) `\ \left( 998 \right)^3 \ `
solution :
( i ) `\ \left( 99 \right)^3 \ `
We have
` \ (99)^3 = ( 100 - 1 )^3\ `
` \ = (100)^3 - (1)^3 - 3 (100) (1) (100 - 1) \ ` ( Using identity VII )
` \= 1000000 - 1 - 300 (100 - 1)\ `
` \= 999999 - 30000 + 300\ `
` \= 969999 + 300\ `
` \= 970299\ `
( ii ) `\ \left( 102 \right)^3 \ `
We have
` \ (102)^3 = ( 100 + 2 )^3\ `
` \ = (100)^3 + (2)^3 + 3 (100) (2) (100 + 2) \ ` ( Using identity VI )
` \= 1000000 + 8 + 600 (100 + 2)\ `
` \= 1000008 + 60000 + 1200\ `
` \= 1060008 + 1200\ `
` \= 1061208\ `
( iii ) `\ \left( 998 \right)^3 \ `
We have
` \ (998)^3 = ( 1000 - 2 )^3\ `
` \ = (1000)^3 - (2)^3 - 3 (1000) (2) (1000 - 2) \ ` ( Using identity VII )
` \= 1000000000 - 8 - 6000 (1000 - 2)\ `
` \= 999999992 - 6000000 + 12000\ `
` \= 993999992 + 12000\ `
` \= 994011992\ `
8. Factorise each of the following:
( i ) ` \ 8 a^3 + b^3 + 12 a^2b + 6 ab^2 \ `
( ii ) ` \ 8a^3 - b^3 - 12a^2b + 6ab^2 \ `
( iii ) ` \ 27 - 125a^3 - 135a + 225a^2 \ `
( iv ) ` \ 64a^3 - 27b^3 - 144a^2b + 108ab^2 \ `
( v ) ` \ 27p^3 - \frac{1} {216} - \frac{9} {2}p^2 + \frac{1} {4}p \ `
solution :
( i ) ` \ 8 a^3 + b^3 + 12 a^2b + 6 ab^2 \ `
The given expression can be written as
` \(2a)^3 + (b)^3 + 3 (4a^2) (b) + 3 ( 2a ) ( b^2 )\ `
` \(2a)^3 + (b)^3 + 3 (2a)^2 (b) + 3 (2a) (b)^2\ `
` \(2a + b )^3\ ` ( Using identity VI )
` \ (2a+b) \ ` ` \ (2a+b) \ ` ` \ (2a+b) \ `
( ii ) ` \ 8a^3 - b^3 - 12a^2b + 6ab^2 \ `
The given expression can be written as
` \(2a)^3 - (b)^3 - 3 (4a^2) (b) + 3 ( 2a ) ( b^2 )\ `
` \(2a)^3 - (b)^3 - 3 (2a) (b) (2a - b)\ `
` \(2a - b )^3\ ` ( Using identity VII )
` \ (2a-b) \ ` ` \ (2a-b) \ ` ` \ (2a-b) \ `
( iii ) ` \ 27 - 125a^3 - 135a + 225a^2 \ `
The given expression can be written as
` \(3)^3 - (5a)^3 - 3 (3^2) (5a) + 3 ( 3 ) ( (5a)^2 )\ `
` \(3)^3 - (5a)^3 - 3 (3) (5a) (3 - 5a) \ `
` \(3 5a )^3\ ` ( Using identity VII )
` \ (3 - 5a) \ ` ` \ (3 - 5a) \ ` ` \ (3-5a) \ `
( iv ) ` \ 64a^3 - 27b^3 - 144a^2b + 108ab^2 \ `
The given expression can be written as
` \(4a)^3 - (3b)^3 - 3 ((4a)^2) (3b) + 3 ( 4a ) ( (3b)^2 )\ `
` \(4a)^3 - (3b)^3 - 3 (4a) (3b) (2a - 3b) \ `
` \(4a - 3b )^3\ ` ( Using identity VII )
` \ (4a-3b) \ ` ` \ (4a-3b) \ ` ` \ (4a-3b) \ `
( v ) ` \ 27p^3 - \frac{1} {216} - \frac{9} {2}p^2 + \frac{1} {4}p \ `
The given expression can be written as
` \(3p)^3 - (\frac {1} {6})^3 - 3 ((3p)^2) (\frac {1} {6}) + 3 ( 3p ) ( (\frac {1} {6})^2 )\ `
` \(3p)^3 - (\frac {1} {6})^3 - 3 (3p) (\frac {1} {6}) (3p - \frac {1} {6}) \ `
` \(3p - \frac {1} {6} )^3\ ` ( Using identity VII )
` \ (3p - \frac{1} {6}) \ ` ` \ (3p - \frac{1} {6}) \ ` ` \ (3p - \frac{1} {6}) \ `
9. Verify :
( i ) ` \ x^3 + y^3 = ( x + y ) \ ` `\ ( x^2 - xy + y^2 ) \ `
( ii ) ` \ x^3 - y^3 = ( x - y ) \ ` ` \( x^2 + xy + y^2 ) \ `
solution :
( i ) ` \ x^3 + y^3 = ( x + y ) \ ` `\ ( x^2 - xy + y^2 ) \ `
We know that,
` \(x+y)^3 = x^3 +y^3 +3xy(x+y) \ `
⇒` \x^3 + y^3 = (x+y)^3 -3xy(x+y)\ `
⇒` \x^3 + y^3 = (x+y) [(x+y)^2 - 3xy]\ ` ( Taking ` \(x+y)\ ` common )
⇒`\x^3+y^3=(x+y)[x^2+y^2+2xy-3xy] \ `
⇒`\x^3+y^3=(x+y)(x^2-xy+y^2) \ `
Proved.
( ii ) ` \ x^3 - y^3 = ( x - y ) \ ` ` \( x^2 + xy + y^2 ) \ `
We know that,
` \(x-y)^3 = x^3 -y^3 -3xy(x-y) \ `
⇒` \x^3 - y^3 = (x-y)^3 +3xy(x-y)\ `
⇒` \x^3 - y^3 = (x-y) [(x-y)^2 + 3xy]\ ` ( Taking ` \(x-y)\ ` common )
⇒`\x^3+-y^3=(x-y)[x^2+y^2-2xy+3xy] \ `
⇒`\x^3-y^3=(x-y)(x^2+xy+y^2) \ `
Proved.
10. Factorise each of the following:
( i ) ` \ 27y^3 + 125z^3 \ `
( ii ) ` \ 64m^3 - 343n^3 \ `
solution :
( i ) ` \ 27y^3 + 125z^3 \ `
The expression, ` \27y^3+125z^3\ ` can be written as` \ (3y)^3+(5z)^3 \ `
` \27y^3+125z^3\ ` = ` \ (3y)^3+(5z)^3 \ `
We know that,
⇒`\x^3+y^3=(x+y)(x^2-xy+y^2) \ `
` \27y^3+125z^3\ ` = ` \ (3y)^3+(5z)^3 \ `
=` \(3y+5z)((3x)^2-(3x)(5z)+(5z)^2) \ `
=` \(3y+5z)(9y^2 -15yz+25z^2)\ `
( ii ) ` \ 64m^3 - 343n^3 \ `
The expression, ` \64m^3-343n^3\ ` can be written as` \ (4m)^3-(7n)^3 \ `
` \64m^3-343n^3\ ` = ` \ (4m)^3+(7n)^3 \ `
We know that,
⇒`\x^3-y^3=(x-y)(x^2+xy+y^2) \ `
` \64m^3-343n^3\ ` = ` \ (4m)^3-(7n)^3 \ `
=` \(4m-7n)((4m)^2+(4m)(7n)+(7n)^2) \ `
=` \(4m-7n)(16m^2 +28mn+49n^2)\ `
11. Factorise : ` \ 27x^3 + y^3 + z^3 - 9xyz \ `
solution :
Factorise : ` \ 27x^3 + y^3 + z^3 - 9xyz \ `
Here, we have
` \ 27x^3 + y^3 + z^3 - 9xyz \ `
` \(3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)\ `
` \(3x + y+z)\ ` ` \[(3x)^2+(y)^2+(z)^2-(3x)(y)-(y)(z)-(z)(3x)]\ `
` \(3x+y+z)\ ` ` \(9x^2+y^2+z^2-3xy-yz-3xz)\ `
12. Verify that `\ x^3 + y^3 + z^3 - 3xyz \ ` ` \ = \frac{1} {2} ( x + y + z ) \ ` ` \ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \ `
solution :
Verify that `\ x^3 + y^3 + z^3 - 3xyz \ ` ` \ = \frac{1} {2} ( x + y + z ) \ ` ` \ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \ `
We know that,
` \x^3+y^3+z^3-3xyz = \ ` ` \(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\ `
⇒`\ x^3 + y^3 + z^3 - 3xyz \ ` ` \ = \frac{1} {2} ( x + y + z ) \ ` ` \ [2 (x^2 + y^2 + z^2-xy-yz-zx) ] \ `
=` \ \frac{1} {2} ( x + y + z ) \ ` ` \ (2x^2 +2y^2 +2z^2 -2xy-2yz-2zx) \ `
=` \ \frac{1} {2} ( x + y + z ) \ ` ` \ [(x^2+y^2-2xy)(y^2+z^2-2yz)(z^2+x^2-2zx)] \ `
=`\ x^3 + y^3 + z^3 - 3xyz \ ` ` \ = \frac{1} {2} ( x + y + z ) \ ` ` \ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \ `
13. If ` \ x + y + z = 0 \ `, show that ` \ x^3 + y^3 + z^3 = 3xyz \ `.
solution :
If ` \ x + y + z = 0 \ `, show that ` \ x^3 + y^3 + z^3 = 3xyz \ `.
We know that,
` \x^3+y^3+z^3-3xyz = \ ` ` \(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\ `
Now, according to the question, let ` \ (x+y+z) = 0, \ `
then, ` \x^3+y^3+z^3 -3xyz = \ ` ` \ (0)(x^2+y^2+z^2–xy–yz–xz) \ `
⇒ ` \x^3+y^3+z^3–3xyz = 0 \ `
⇒ ` \x^3+y^3+z^3=3xyz \ `
14. Without actually calculating the cubes, find the value of each of the following:
( i ) ` \ ( -12 )^3 + ( 7 )^3 + ( 5 )^3 \ `
( ii ) ` \ ( 28 )^3 + ( -15 )^3 + ( -13 )^3 \ `
solution :
( i ) ` \ ( -12 )^3 + ( 7 )^3 + ( 5 )^3 \ `
Let a ` \ = −12 \ `
` \b = 7\ `
` \c = 5\ `
We know that if ` \x+y+z = 0,\ ` then ` \x^3+y^3+z^3=3xyz \ `.
Here, ` \−12+7+5=0\ `
` \(−12)^3+(7)^3+(5)^3 = 3xyz\ `
` \= 3×(-12)×7×5\ `
` \= -1260\ `
( ii ) ` \ ( 28 )^3 + ( -15 )^3 + ( -13 )^3 \ `
Let a ` \ = 28 \ `
` \b = -15\ `
` \c = -13\ `
We know that if ` \x+y+z = 0,\ ` then ` \x^3+y^3+z^3=3xyz \ `.
Here, ` \28+(-15)+(-13)=0\ `
` \(28)^3+(-15)^3+(-13)^3 = 3xyz\ `
` \= 3×(28)×(-15)×(-13)\ `
` \= 16380\ `
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
( i ) ` \ Area : 25a^2 - 35a + 12 \ `
( ii ) ` \ Area : 35y^2 - 13y - 12 \ `
solution :
( i ) ` \ Area : 25a^2 - 35a + 12 \ `
By splitting method:
If we can find two numbers p and q such that,
` \p+q=-35\ ` and p q = 25 × 12 = 300
then we can get the factor.
So, let us look for the pair of factors of 300.
Some are − 15 and − 20.
of these pairs, − 15 and − 20 will giue us p + q = − 35.
So, ` \ 25a^2-35a+12\ ` ` \ = 25a^2 +[(-15)+(-20)]a+12\ `
= ` \25a^2-15a-20a+12\ `
= ` \5a(5a-3)-4(5a-3)\ `
= ` \(5a-4) (5a-3)\ `
= Possible expression for length ` \= (5a–4)\ `
= Possible expression for breadth ` \= (5a –3)\ `
( ii ) ` \ Area : 35y^2 - 13y - 12 \ `
By splitting method:
If we can find two numbers p and q such that,
` \p+q=-13\ ` and p q = 35 × − 12 = −420
then we can get the factor.
So, let us look for the pair of factors of − 420.
Some are − 15 and 28.
of these pairs, − 15 and 28 will giue us p + q = − 13.
So, ` \ 35y^2-13y-12\ ` ` \ = 35y^2 +[(-15)+28]y-12\ `
= ` \35y^2-15y+28y-12\ `
= ` \5y(7y-3)-4(7y-3)\ `
= ` \(5y-4) (7y-3)\ `
= Possible expression for length ` \= (5y–4)\ `
= Possible expression for breadth ` \= (7y –3)\ `
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
( i ) ` \ Volume : 3x^2 - 12x \ `
( ii ) ` \ Volume : 12ky^2 + 8ky - 20k \ `
solution :
( i ) ` \ Volume : 3x^2 - 12x \ `
` \3x^2–12x\ ` can be written as ` \3x(x–4)\ `
Taking 3x common both the terms.
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x – 4)
( ii ) ` \ Volume : 12ky^2 + 8ky - 20k \ `
` \12ky^2+8ky–20k\ ` can be written as ` \4k(3y^2+2y–5)\ `
Taking 4k common both the terms.
Here, ` \4k(3y^2+2y-5)\ `
By splitting method:
` \4k (3y^2+2y-5)\ ` is written as ` \4k (3y^2+5y-3y-5)\ `
` \4k [ y(3y+5)-1(3y+5)]\ `
` \4k [(3y+5) (y-1)]\ `
` \4k (3y+5) (y-1)\ `
Possible expression for length = 4k
Possible expression for breadth = ( 3y + 5 )
Possible expression for height = (y – 1)
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