Advertisements
Advertisements
Question
Factorise:
a3 – 8b3 – 64c3 – 24abc
Advertisements
Solution
a3 – 8b3 – 64c3 – 24abc = (a)3 + (–2b)3 + (–4c)3 – 3 × (a) × (–2b) × (–4c)
= (a – 2b – 4c)[(a)2 + (–2b)2 + (–4c)2 – a(–2b) – (–2b)(–4c) – (–4c)(a)] ...[Using identity, a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)]
= (a – 2b – 4c)(a2 + 4b2 + 16c2 + 2ab – 8bc + 4ac)
APPEARS IN
RELATED QUESTIONS
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = kx2 – 3x + k
Factorise:
2x2 + 7x + 3
Factorise:
x3 – 2x2 – x + 2
Find the factor of the polynomial given below.
`1/2x^2 - 3x + 4`
Factorize the following polynomial.
(y2 + 5y) (y2 + 5y – 2) – 24
One of the factors of (25x2 – 1) + (1 + 5x)2 is ______.
Factorise:
84 – 2r – 2r2
Factorise:
1 + 64x3
Factorise:
`2sqrt(2)a^3 + 8b^3 - 27c^3 + 18sqrt(2)abc`
If both x – 2 and `x - 1/2` are factors of px2 + 5x + r, show that p = r.
