(i) a4 + b4 – 7a2b2
(ii) x4 – 14x2 + 1
Answer :
(i) a4 + b4 – 7a2b2
Above terms can be written as,
a4 + b4 + 2a2b2 – 9a2b2
We know that, (a + b)2 = a2 + 2ab + b2,
[(a2)2 + (b2)2 + (2 × a2 × b2)] – (3ab)2
(a2 + b2)2 – (3ab)2
We know that, a2 – b2 = (a + b) (a – b)
(a2 + b2 + 3ab) (a2 + b2 – 3ab)
(ii) x4 – 14x2 + 1
x4 – 14x2 + 1
Above terms can be written as,
x4 + 2x2 + 1 – 16x2
We know that, (a + b)2 = a2 + 2ab + b2,
So, [(x2)2 + (2 × x2 × 1) + 12] – 16x2
(x2 + 1)2 – (4x)2
We know that, a2 – b2 = (a + b) (a – b)
(x2 + 1 + 4x) (x2 + 1 – 4x)
More Solutions: