Question

Mathematics:
1) If $\bf{\frac{{a}^{3}+{3ab}^{2}}{{3a}^{2}b+{b}^{3}}=\frac{{c}^{3}+{3cd}^{2}}{{3c}^{2}d+{d}^{3}}}$, then show that a, b, c and d are in proportion.

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Answer 1

${\underline{\underline{\bf{Required\:solution:}}}}$

$\sf{Given,}$

$\large\underline{\boxed{\sf{\frac{{a}^{3}+{3ab}^{2}}{{3a}^{2}b+{b}^{3}}=\frac{{c}^{3}+{3cd}^{2}}{{3c}^{2}d+{d}^{3}}}}}$

$\sf{Applying\:componendo\:and\:dividendo,}$

$\implies\large{\sf{ \frac{( {a}^{3} + {3ab}^{2}) + ( {3a}^{2} b + {b}^{3} )}{( {a}^{3} + {3ab}^{2}) - ( {3a}^{2} b + {b}^{3} )} = \frac{( {c}^{3} + {3cd}^{2}) + ( {3c}^{2} d + {d}^{3} )}{( {c}^{3} + {3cd}^{2}) - ( {3c}^{2} d + {d}^{3})} }}$

$\implies\large{\sf{ \frac{ {a}^{3} + {b}^{3} + 3ab(a + b) }{ {a}^{3} - {b}^{3} - 3ab(a - b) } = \frac{ {c}^{3} + {d}^{3} + 3cd(c + d) }{ {c}^{3} - {d}^{3} - 3cd(c - d) } }}$

$\implies\large{\sf{ \frac{ {(a + b)}^{3} }{ {(a - b)}^{3} } = \frac{ {(c + d)}^{3} }{ {(c - d)}^{3} } }}$

$\implies\large{\sf{ \frac{a + b}{a - b} = \frac{c + d}{c - d} }}$

$\sf{Again,\: applying\:componendo\:and\:dividendo,}$

$\implies\large{\sf{ \frac{(a + b) + (a - b)}{(a + b) - (a - b)} = \frac{(c + d) + (c - d)}{(c + d) - (c - d)} }}$

$\implies\large{\sf{ \frac{2a}{2b} = \frac{2c}{2d}}}$

$\implies\underline{\boxed{\sf{\red{ \frac{a}{b} = \frac{c}{d} }}}}$

$\:\:$

$\therefore\underline{\sf{a,\:b,\:c\:and\:d\:\:are\:in\: proportion.}}$

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