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

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Hint: Here we will prove the equation by simplifying the LHS and getting the result as the RHS through the given condition.

Complete step-by-step answer:
You have to prove If $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}$ then $\dfrac{{2{a^4}{b^2} + 3{a^2}{e^2} - 5{e^4}f}}{{2{b^6} + 3{b^2}{f^2} - 5{f^5}}} = \dfrac{{{a^4}}}{{{b^4}}}$.
Taking L.H.S we have
$\dfrac{{2{a^4}{b^2} + 3{a^2}{e^2} - 5{e^4}f}}{{2{b^6} + 3{b^2}{f^2} - 5{f^5}}}$.
Taking ${a^4}$common from Numerator and ${b^4}$from the denominator.
$ \Rightarrow \dfrac{{{a^4}(2{b^2} + 3\dfrac{{{e^2}}}{{{a^2}}} - 5\dfrac{{{e^4}}}{{{a^4}}}f)}}{{{b^4}(2{b^2} + 3\dfrac{{{f^2}}}{{{b^2}}} - 5\dfrac{{{f^5}}}{{{b^4}}})}} \to (1)$
Since we have \[\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} \Rightarrow \dfrac{a}{b} = \dfrac{e}{f} \Rightarrow \dfrac{e}{a} = \dfrac{f}{b}\]
Put this value in numerator of equation 1
$ \Rightarrow \dfrac{{{a^4}(2{b^2} + 3\dfrac{{{e^2}}}{{{a^2}}} - 5\dfrac{{{e^4}}}{{{a^4}}}f)}}{{{b^4}(2{b^2} + 3\dfrac{{{f^2}}}{{{b^2}}} - 5\dfrac{{{f^5}}}{{{b^4}}})}} = \dfrac{{{a^4}(2{b^2} + 3\dfrac{{{f^2}}}{{{b^2}}} - 5\dfrac{{{f^5}}}{{{b^4}}})}}{{{b^4}(2{b^2} + 3\dfrac{{{f^2}}}{{{b^2}}} - 5\dfrac{{{f^5}}}{{{b^4}}})}} = \dfrac{{{a^4}}}{{{b^4}}} = R.H.S$
Hence, L.H.S=R.H.S.

Note: In this type of question what you have to prove takes common from Numerator and denominator and then simplify according to given condition, you will get your answer.