Chapter 2 Polynomials - Exercise 2.5 | Set 1
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 โ y3 = (x - y)3 + 3xy(x - y)
x3โ y3 = (x - y) [(x - y)2 + 3xy]
x3 โ y3 = (x - y) [(x2 + y2 โ 2xy) + 3xy]
Therefore, x3 + y3 = (x โ 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 โ xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) [(3y)2 โ (3y)(5z) + (5z)2]
= (3y + 5z) (9y2 โ 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 โ 9xyz
Solution:
27x3 + y3 + z3 - 9xyz can also be written as (3x)3 + y3 + z3 - 3(3x)(y)(z)
27x3 + y3 + z3 โ 9xyz = (3x)3 + y3 + z3 โ 3(3x)(y)(z)
Formula, x3 + y3 + z3 โ 3xyz = (x + y + z) (x2 + y2 + z2 โ xy โ yz โ zx)
27x3 + y3 + z3 โ 9xyz = (3x)3 + y3 + z3 โ 3(3x)(y)(z)
= (3x + y + z) [(3x)2 + y2 + z2 โ 3xy โ yz โ 3xz]
= (3x + y + z) (9x2 + y2 + z2 โ 3xy โ yz โ 3xz)
Question 12. Verify that: x3 + y3 + z3 โ 3xyz = (1/2) (x + y + z) [(x โ y)2 + (y โ z)2 + (z โ x)2]
Solution:
Formula, x3 + y3 + z3 โ 3xyz = (x + y + z)(x2 + y2 + z2 โ xy โ yz โ xz)
Multiplying by 2 and dividing by 2
= (1/2) (x + y + z) [2(x2 + y2 + z2 โ xy โ yz โ xz)]
= (1/2) (x + y + z) (2x2 + 2y2 + 2z2 โ 2xy โ 2yz โ 2xz)
= (1/2) (x + y + z) [(x2 + y2 โ 2xy) + (y2 + z2 โ 2yz) + (x2 + z2 โ 2xz)]
= (1/2) (x + y + z) [(x โ y)2 + (y โ z)2 + (z โ x)2]
Therefore, x3 + y3 + z3 โ 3xyz = (1/2) (x + y + z) [(x โ y)2 + (y โ z)2 + (z โ 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 โ xy โ yz โ xz)
Given, (x + y + z) = 0,
Then, x3 + y3 + z3 - 3xyz = (0) (x2 + y2 + z2 โ xy โ yz โ xz)
x3 + y3 + z3 โ 3xyz = 0
Therefore, x3 + y3 + z3 = 3xyz
Question 14. Without actually calculating the cubes, find the value of each of the following:
(i) (โ12)3 + (7)3 + (5)3
Solution:
Let,
x = โ12
y = 7
z = 5
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have โ12 + 7 + 5 = 0
Therefore, (โ12)3 + (7)3 + (5)3 = 3xyz
= 3 ร -12 ร 7 ร 5
= -1260
(ii) (28)3 + (โ15)3 + (โ13)3
Solution:
Let,
x = 28
y = โ15
z = โ13
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have, x + y + z = 28 โ 15 โ 13 = 0
Therefore, (28)3 + (โ15)3 + (โ13)3 = 3xyz
= 3 (28) (โ15) (โ13)
= 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 โ 35a + 12
Solution:
Using splitting the middle term method,
25a2 โ 35a + 12
25a2 โ 35a + 12 = 25a2 โ 15a โ 20a + 12
= 5a(5a โ 3) โ 4(5a โ 3)
= (5a โ 4) (5a โ 3)
Possible expression for length & breadth is = (5a โ 4) & (5a โ 3)
(ii) Area : 35y2 + 13y โ 12
Solution:
Using the splitting the middle term method,
35y2 + 13y โ 12 = 35y2 โ 15y + 28y โ 12
= 5y(7y โ 3) + 4(7y โ 3)
= (5y + 4) (7y โ 3)
Possible expression for length & breadth is = (5y + 4) & (7y โ 3)
Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x2 โ 12x
Solution:
3x2 โ 12x can also be written as 3x(x โ 4)
= (3) (x) (x โ 4)
Possible expression for length, breadth & height = 3, x & (x - 4)
(ii) Volume: 12ky2 + 8ky โ 20k
Solution:
12ky2 + 8ky โ 20k can also be written as 4k (3y2 + 2y โ 5)
12ky2 + 8kyโ 20k = 4k(3y2 + 2y โ 5)
Using splitting the middle term method.
= 4k (3y2 + 5y โ 3y โ 5)
= 4k [y(3y + 5) โ 1(3y + 5)]
= 4k (3y + 5) (y โ 1)
Possible expression for length, breadth & height= 4k, (3y + 5) & (y - 1)
