If $ f(x)=x^{2} $ and $ g(x)=x^{3}, $ then $ \frac{f(b)-f(a)}{g(b)-g(a)}= $
Given:
\( f(x)=x^{2} \) and \( g(x)=x^{3} \)
To do:
We have to find the value of \( \frac{f(b)-f(a)}{g(b)-g(a)} \).
Solution:
We know that,
$a^3-b^3=(a-b)(a^2+ab+b^2)$
$a^2-b^2=(a-b)(a+b)$
Therefore,
$f(b)=b^2, f(a)=a^2, g(b)=b^3, g(a)=a^3$
$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{b^2-a^2}{b^3-a^3}$
$=\frac{(b-a)(b+a)}{(b-a)(b^2+ab+a^2)}$
$=\frac{a+b}{a^2+ab+b^2}$
Hence, $\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{a+b}{a^2+ab+b^2}$.
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