Question

Q1. If a/b = c/d , show that:
• a³c + ac³ / b³+bd³ = (a+c)⁴/(b+d)⁴​

Answer 1

$\Large{\bold{\underline{\underline \mathfrak{Appropriate\: question:-}}}}$

. If a/b = c/d , show that:

• a³c + ac³ / b³d+bd³ = (a+c)⁴/(b+d)⁴

$\Large{\bold{\underline{\underline \mathfrak{given \: information:-}}}}$

• $\bf{} \dfrac{a}{b} = \dfrac{c}{d}$

$\Large{\bold{\underline{\underline \mathfrak{Need \: to \: find?}}}}$

• We have to show that both the sides R.H.S. and L.H.S. are equal.

$\Large{\bold{\underline{\underline \mathfrak{Step \: by \: step \: explaination:-}}}}$

• Let us put the given ratios equal to k
• Then obtain anticident of every ratio in terms of k.
• After that substitute the values in the terms of k with that after doing simplify it.

Let's solve it.

• $\hookrightarrow \: \bf{ \dfrac{a {}^{3} + ac {}^{3} }{b {}^{3} d \: + \: bd {}^{3} } = \dfrac{(a + c) {}^{4} }{(b + d) {}^{4} } }$

Calculation for L.H.S. (Left hand side)

We know that a/b = c/d as c/d is equal to k thus a/b would also be k.

• By cross multiplying calculating value of a and c.
• $\text{a = bk \: and \: c = dk}$

Substituting values.

• $\hookrightarrow \: \bf{ \dfrac{bk {}^{3} \times dk + bk \times dk {}^{3} }{b {}^{3}d + bd {}^{3} } }$

Solving.

• $\hookrightarrow \: \bf{ \dfrac{b {}^{3} d {}^{4} k {}^{4} \: + \: bd {}^{3} k {}^{4} }{b {}^{3}d \: + \: bd {}^{3} } }$

• $\hookrightarrow \: \bf{ \dfrac{k {}^{4} (b {}^{3} d \: + \: bd {}^{3} )}{(b {}^{3}d + bd {}^{3} ) } }$

Cancelling them.

• $\hookrightarrow \: \bf{ \dfrac{k {}^{4} \: (\cancel{b {}^{3} d \: + \: bd {}^{3} })}{( \cancel{b {}^{3}d + bd {}^{3} )} } }$

Now,

• $\hookrightarrow \: \boxed{ \underline{\bf{k {}^{4}} }}$

__________

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Answer 2

$\boxed{\boxed{\huge{\blue{\bold{\underline{Answer}}}}}}$

$Let \: \frac{a}{b} = \frac{c}{d} = k</p><p> \\ \\ = > a = bk \: and \: c = dk$

$L..H.S = \frac{a {}^{3}c \: + \: ac {}^{3} }{b {}^{3}d \: + \: bd {}^{3} }$

$= \frac{ac(a {}^{2} \: + \: c {}^{2}) }{bd(b { }^{2} \: + \: d {}^{2} )}$

$= \frac{(bk \: \times \: dk)(b {}^{2} k {}^{2} \: + \: d {}^{2}k {}^{2} }{bd(b {}^{2} \: + \: d {}^{2} )}$

$= \frac{k {}^{2} \: \times \: k {}^{2} (b {}^{2} \: + \: d {}^{2} ) }{b {}^{2} \: + \: d {}^{2} }$

$= k {}^{4}$

$R.H.S = \frac{(a \: + \: c) {}^{4} }{(b \: + \: d) {}^{4} } = \frac{(bk \: + \: dk) {}^{4} }{(b \: + \: d) {}^{4} } \\ \\ \: = [ \frac{k(b \: + \: d)}{b \: + \: d} ] {}^{4} = k {}^{4}$

$Hence, = \frac{a {}^{3} c \: + \: ac {}^{3} } {b {}^{3}d \: + \: bd {}^{3} } = \frac{(a \: + \: c) {}^{4} }{(b \: + \: d) {}^{4} }$