\[\frac{a^4-b^2}{a-b}\]
\[\\frac{\(a^\2\-b)\(a^\2\+b)\}\{a-b}\] i factor the numerator but i don't know what you want to do with this expression
\[\frac{a^4-b^2}{a-b}=?\] a)a^2+b^2 b) (a+b)(a^2-b^2) c)(a+b)(a^2+b^2)
let's try multiplying the conjugate of the bottom to top and bottom
\[\frac{(a^2-b)(a^2+b)}{a-b} \cdot \frac{a+b}{a+b}=\frac{(a^2-b)(a^2+b)(a+b)}{a^2-b^2}\] nope we can't simplify to anythose
if a=2 and b=3 we have \[\frac{2^4-3^2}{2-3}=-7\] \[2^2+3^2=13\] \[(2+3)(2^2-3^2)=5(4-9)=5(-5)=-25\] \[(2+3)(2^2+3^2)=5(4+9)=5(13)=65\]
see we don;t get the same for any of them
oh sorry myininaya the question was wrong it is saying the question as a^-b^4/a-b but it was given in the question a^-b^2/a-b ...... so don't waste ur time because of this question thanks again
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