Question

Factorise: 125p^(3) - 729q^(3)​

Answer 1

Step-by-step explanation:

Given:-

125p^(3) - 729q^(3)

To find:-

Factorise: 125p^(3) - 729q^(3)

Solution:-

Given expression is 125p^(3) - 729q^(3)

125 = 5×5×5 = 5^3

729=9×9×9=9^3

125p^(3) - 729q^(3)

=>5^3 p^3 - 9^3 q^3

We know that

a^m × b^m = (ab)^m

=> (5p)^3 - (9q)^3

It is in the form of a^3-b^3

Where , a= 5p and b = 9q

We know that

a^3-b^3 = (a-b)(a^2+ab+b^2)

=> (5p-9q)[(5p)^2+(5p)(9q)+(9q)^2]

=>(5p-9q)(25p^2+45pq+81q^2)

125p^(3)-729q^(3)=(5p-9q)(25p^2+45pq+81q^2)

Answer:-

Factorization for the given problem is

(5p-9q)(25p^2+45pq+81q^2)

Used formulae:-

• a^m × b^m = (ab)^m

• a^3-b^3 = (a-b)(a^2+ab+b^2)

Answer 2

Answer:

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