Question 9. Verify:
(i) x3 + y3 = (x + y) (x2 – xy + y2)
Solution:
Formula (x + y)3 = x3 + y3 + 3xy(x + y)
x3 + y3 = (x + y)3 – 3xy(x + y)
x3 + y3 = (x + y) [(x + y)2 – 3xy]
x3 + y3 = (x + y) [(x2 + y2 + 2xy) – 3xy]
Therefore, x3 + y3 = (x + y) (x2 + y2 – xy)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
Formula, (x – y)3 = x3 – y3 – 3xy(x – y)
x3 − y3 = (x – y)3 + 3xy(x – y)
x3− y3 = (x – y) [(x – y)2 + 3xy]
x3 − y3 = (x – y) [(x2 + y2 – 2xy) + 3xy]
Therefore, x3 + y3 = (x – y) (x2 + y2 + xy)
Question 10. Factorize each of the following:
(i) 27y3 + 125z3
Solution:
27y3 + 125z3 can also be written as (3y)3 + (5z)3
27y3 + 125z3 = (3y)3 + (5z)3
Formula x3 + y3 = (x + y) (x2 – xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z) (9y2 – 15yz + 25z2)
(ii) 64m3 – 343n3
Solution:
64m3 – 343n3 can also be written as (4m)3 – (7n)3
64m3 – 343n3 = (4m)3 – (7n)3
Formula x3 – y3 = (x – y) (x2 + xy + y2)
64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n) [(4m)2 + (4m)(7n) + (7n)2]
Question 11. Factorise: 27x3 + y3 + z3 – 9xyz
Solution:
27x3 + y3 + z3 – 9xyz can also be written as (3x)3 + y3 + z3 – 3(3x)(y)(z)
27x3 + y3 + z3 – 9xyz = (3x)3 + y3 + z3 – 3(3x)(y)(z)
Formula, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
27x3 + y3 + z3 – 9xyz = (3x)3 + y3 + z3 – 3(3x)(y)(z)
= (3x + y + z) [(3x)2 + y2 + z2 – 3xy – yz – 3xz]
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)
Question 12. Verify that: x3 + y3 + z3 – 3xyz = (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Solution:
Formula, x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
Multiplying by 2 and dividing by 2
= (1/2) (x + y + z) [2(x2 + y2 + z2 – xy – yz – xz)]
= (1/2) (x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2xz)
= (1/2) (x + y + z) [(x2 + y2 − 2xy) + (y2 + z2 – 2yz) + (x2 + z2 – 2xz)]
= (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Therefore, x3 + y3 + z3 – 3xyz = (1/2) (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Solution:
Formula, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – xz)
Given, (x + y + z) = 0,
Then, x3 + y3 + z3 – 3xyz = (0) (x2 + y2 + z2 – xy – yz – xz)
x3 + y3 + z3 – 3xyz = 0
Therefore, x3 + y3 + z3 = 3xyz
Question 14. Without actually calculating the cubes, find the value of each of the following:
(i) (−12)3 + (7)3 + (5)3
Solution:
Let,
x = −12
y = 7
z = 5
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have −12 + 7 + 5 = 0
Therefore, (−12)3 + (7)3 + (5)3 = 3xyz
= 3 × -12 × 7 × 5
= -1260
(ii) (28)3 + (−15)3 + (−13)3
Solution:
Let,
x = 28
y = −15
z = −13
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have, x + y + z = 28 – 15 – 13 = 0
Therefore, (28)3 + (−15)3 + (−13)3 = 3xyz
= 3 (28) (−15) (−13)
= 16380
Question 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a2 – 35a + 12
Solution:
Using splitting the middle term method,
25a2 – 35a + 12
25a2 – 35a + 12 = 25a2 – 15a − 20a + 12
= 5a(5a – 3) – 4(5a – 3)
= (5a – 4) (5a – 3)
Possible expression for length & breadth is = (5a – 4) & (5a – 3)
(ii) Area : 35y2 + 13y – 12
Solution:
Using the splitting the middle term method,
35y2 + 13y – 12 = 35y2 – 15y + 28y – 12
= 5y(7y – 3) + 4(7y – 3)
= (5y + 4) (7y – 3)
Possible expression for length & breadth is = (5y + 4) & (7y – 3)
Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x2 – 12x
Solution:
3x2 – 12x can also be written as 3x(x – 4)
= (3) (x) (x – 4)
Possible expression for length, breadth & height = 3, x & (x – 4)
(ii) Volume: 12ky2 + 8ky – 20k
Solution:
12ky2 + 8ky – 20k can also be written as 4k (3y2 + 2y – 5)
12ky2 + 8ky– 20k = 4k(3y2 + 2y – 5)
Using splitting the middle term method.
= 4k (3y2 + 5y – 3y – 5)
= 4k [y(3y + 5) – 1(3y + 5)]
= 4k (3y + 5) (y – 1)
Possible expression for length, breadth & height= 4k, (3y + 5) & (y – 1)
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Last Updated :
05 Apr, 2021
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