Question

a^4+b^4-7a^2b^2 factorise​

Answer 1

$\textbf{Given:}$

$a^4+b^4-7\,a^2b^2$

$=(a^2)^2+(b^2)^2-7\,a^2b^2$

$\text{Using the following identity}$

$\boxed{\bf(x+y)^2=x^2+y^2+2xy\implies\,x^2+y^2=(x+y)^2-2xy}$

$=(a^2+b^2)^2-2\,a^2b^2-7\,a^2b^2$

$=(a^2+b^2)^2-9\,a^2b^2$

$=(a^2+b^2)^2-(3\,ab)^2$

$\text{Using the following identity}$

$\boxed{\bf\,x^2-y^2=(x+y)(x-y)}$

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

$\therefore\boxed{\bf\,a^4+b^4-7\,a^2b^2=(a^2+b^2+3ab)(a^2+b^2-3ab)}$

Find more:

1.X2-(a-1/a)x+1 factorise the following

https://brainly.in/question/4301404#

2.Factorise x^4-(x-z)^4​

https://brainly.in/question/14993176

Answer 2

Step-by-step explanation:

a^4 + b^4 - 7a^2b^2

(a^2)^2 + (b^2)^2 - 7a^2b^2

(a^2 + b^2)^2 - 2a^2b^2 - 7a^2b^2

(a^2 + b^2)^2 - 9a^2b^2

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

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

a^4 + b^4 - 7a^2b^2

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