Question

7
6) A4 + b4 - 7a2b2 factorise​

Answer 1

$\textbf{Given:}$

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

$\textbf{To find:}$

$\text{Factors of a^4+b^4-7\,a^2b^2 }$

$\textbf{Solution:}$

$\text{Consider,}$

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

$\text{Using the following identity}$

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

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

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

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

$\text{Using the following identity}$

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

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

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

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Answer 2

$\red{a^{4} + b^{4} - 7a^{2}b^{2} }\\= (a^{2})^{2} + </p><p> (b^{2})^{2} +2\times a^{2} \times b^{2} -2a^{2}b^{2}-7a^{2}b^{2}$

$= (a^{2}+b^{2})^{2} - 9a^{2}b^{2}$

$\boxed { \pink { Since, x^{2} + y^{2}+2xy = (x+y)^{2} }}$

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

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

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

$= (a^{2}+b^{2}+3ab )[(a-b)^{2} - ab ]$

$= (a^{2}+b^{2}+3ab )[(a-b)^{2} - (\sqrt{ab})^{2}]$

$= (a^{2}+b^{2}+3ab )(a-b+\sqrt{ab})(a-b-\sqrt{ab})$

Therefore.,

$\red{a^{4} + b^{4} - 7a^{2}b^{2} }\\\green {= (a^{2}+b^{2}+3ab )(a-b+\sqrt{ab})(a-b-\sqrt{ab})}$

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