Advertisements
Advertisements
Question
Factorise:
`16a^3 - 128/b^3`
Sum
Advertisements
Solution
It is known that,
a3 − b3 = (a − b)(a2 + ab + b2)
`16a^3 - 128/b^3`
= `16[a^3 - 8/b^3]`
= `16[(a)^3 - (2/b)^3]`
= `16[(a - 2/b) {(a)^2 + (a) xx (2/b) + (2/b)^2}]`
= `16(a - 2/b) (a^2 + (2a)/b + 4/b^2)`
shaalaa.com
Is there an error in this question or solution?
RELATED QUESTIONS
Simplify:
\[\frac{4 x^2 - 11x + 6}{16 x^2 - 9}\]
Factorise:
x3 − 64y3
Factorise:
27m3 − 216n3
Factorise:
343a3 − 512b3
Simplify:
(a + b)3 − a3 − b3
Simplify:
(3xy − 2ab)3 − (3xy + 2ab)3
Factorise: 27p3 - 125q3.
Factorise: 54p3 - 250q3.
Simplify: (a - b)3 - (a3 - b3)
Factorise the following:
27x3 – 8y3
