Advertisements
Advertisements
प्रश्न
Factorise:
a3 – 8b3 – 64c3 – 24abc
Advertisements
उत्तर
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
संबंधित प्रश्न
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:
p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = `2x^2+kx+sqrt2`
Factorise:
6x2 + 5x – 6
Factorize the following polynomial.
(x2 – 6x)2 – 8 (x2 – 6x + 8) – 64
Factorize the following polynomial.
(y2 + 5y) (y2 + 5y – 2) – 24
Factorize the following polynomial.
(x – 3) (x – 4)2 (x – 5) – 6
One of the factors of (25x2 – 1) + (1 + 5x)2 is ______.
Show that p – 1 is a factor of p10 – 1 and also of p11 – 1.
Factorise the following:
1 – 64a3 – 12a + 48a2
Find the following product:
(2x – y + 3z)(4x2 + y2 + 9z2 + 2xy + 3yz – 6xz)
