Question

10. Factorise each of the following:
(i) 27y^3 + 125z^3​

Answer 1

We have to factorize -

27 y^{3}+125 z^{3}=(3 y)^{3}+(5 z)^{3} \rightarrow(1) [ Since 27 = cube of 3, 125 = cube of 5]

We know that a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)

Let, a = 3y, b= 5z

Hence,

(3 y)^{3}+(5 z)^{3}=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right]

(3 y)^{3}+(5 z)^{3}=(3 y+5 z)\left(9 y^{2}+25 z^{2}-15 y z\right) \rightarrow(2)

From eq. (1) and (2) we get,

27 y^{3}+125 z^{3} \equiv(3 y+5 z)\left(9 y^{2}+25 z^{2}-15 y z\right)

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Answer 2

Answer:

Step-by-step explanation:this can also be written as

(3y)³+(5z)³.............1

Identity a³+b³=(a+b)(a²-ab+b²)

a=3y

b=5z

(3y+5z)(9y²-15yz+25z²)

Which is equal to..............1