Question

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f},$ prove that $\frac{2a^4{b}^2+3a^2{e}^2-5e^{4}f}{2b^6+3b^2{f}^2-5f^5}=\frac{a^4}{b^4}$

Answer 1

Solution

To Prove

$\frac{2a^4{b}^2+3a^2{e}^2-5e^{4}f}{2b^6+3b^2{f}^2-5f^5}=\frac{a^4}{b^4}$

Taking LHS , We have

$=\frac{2a^4{b}^2+3a^2{e}^2-5e^{4}f}{2b^6+3b^2{f}^2-5f^5}$

Taking $a^4$ Common from Numerator and $b^4$ from the Denominator .

$=\frac{a^4[2b^2+\frac{3e^2}{a^2}-\frac{5e^{4}f}{a^4}]}{b^4[2b^2+\frac{3f^2}{b^2}-\frac{5f^5}{b^4}]}$

Since , We have

$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$

From this , We have

$\frac{e}{a}=\frac{f}{b}$

Therefore In Above Equation On putting value , we have

$=\frac{a^4[2b^2+\frac{3f^2}{b^2}-\frac{5f^{4}f}{b^4}]}{b^4[2b^2+\frac{3f^2}{b^2}-\frac{5f^5}{b^4}]}$

$=\frac{a^4[2b^2+\frac{3f^2}{b^2}-\frac{5f^5}{b^4}]}{b^4[2b^2+\frac{3f^2}{b^2}-\frac{5f^5}{b^4}]}$

$=\frac{a^4}{b^4}$

$=RHS$

Hence , $LHS=RHS$