If ab - cd - ef - show that a3 b - 2c2 e - 3 a e2 fb4 - 2d2 f - 3 bf3 - acebdf -

Let ab = cd = ef #160;= k ; then a = bk, c = dk, e =fk ; ∴a^3 b + 2c^2 e 3 a e^2 fb^4 + 2d^2 f 3 bf^3 #160; = b^4 k^3 + 2 d^2 fk^3 3bf^ ...

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Laws of Exponents for Real Numbers

If ab = cd ...

Question

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

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

\frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}} = \frac{a c e}{b d f}

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

(b d f) . {(\frac{a + b}{a} + \frac{c + d}{d} + \frac{e + f}{f})}^{3} = 27 (a + b) (c + d) (e + f)

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

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

A F = \frac{1}{3} A C .

a^{3} b + a b^{3} < a^{4} + b^{4}

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