Question

prove this
if a:b::c:d
(a+c)^3:(b+d)^3::a(a-c)^2:b(b-d)^2

Answer 1

We start with the proportion a:b=c:da:b=c:d. Each proportion has something called an alternate form, and if the proportion is true, then so is the alternate form. The alternate form of the proportion a:b=c:da:b=c:d is a:c=b:da:c=b:d. This proportion says two ratios are equal. Call that ratio rr.

One of the operations that you can perform on ratios is compounding one with another which is essentially one ratio by the other. If you compound a ratio with itself, then you're squaring it. Do that with the ratios in a:c=b:da:c=b:d conclude a2:c2=b2:d2a2:c2=b2:d2. This ratio is r2r2.

If you have a proportion, like w:x=y:zw:x=y:z you can add the antecedent and consequent of the second ratio the those of the first to get an equal ratio, that is (w+y):(x+z)=w:x=y:z(w+y):(x+z)=w:x=y:z. Do that with the proportion a2:c2=b2:d2a2:c2=b2:d2 to conclude that

(a2+b2):(c2+d2)=a2:c2=b2:d2(a2+b2):(c2+d2)=a2:c2=b2:d2.

In other words,

a2+b2c2+d2a2+b2c2+d2

is equal to r2r2, the square of the ratio

r=ac=bdr=ac=bd.

Someone who was familiar with these rules of proportions would have seen right away the three steps to make the conclusion. That was all before the invention of symbolic algebra. We've had symbolic algebra for 500 years now, and the rules of proportions have receded into history

Answer 2

Hope u like my process
=====================
Given
=-=-=-=
$= > \bf a : b : : c : d \\ \\ \: \: \: so \: \\ \\ = > \boxed{ \bf \frac{a}{b} = \frac{c}{d} } \: \: ...... \: \: (1) \\ \\ \: \: \: \: \underline{ \: \: now \: \: } \\ \\ = > \frac{a}{b} = \frac{c}{d} \\ \\ \underline{ \: \: \: \bf multiplying \: \: both \: \: sides \: \: by \: \: \frac{b}{c} } \\ \\ = > \frac{ a }{c} = \frac{b}{d} \\ \\ \underline{ \: \: \: \bf subtracting \: \: 1 \: \: on \: \: both \: \: sides} \\ \\ = > \frac{a}{c} - 1 = \frac{b}{d} - 1 \\ \\ = > \frac{a - c}{c} = \frac{b - d}{d} \\ \\ \underline{ \bf\: \: \: squaring \: \: both \: \: sides} \\ \\ = > {( \frac{a - c}{c} )}^{2} = {( \frac{b - d}{d} )}^{2} \\ \\ = > \frac{ {(a - c)}^{2} }{ {c}^{2} } = \frac{ {(b - d)}^{2} }{ {d}^{2} } \\ \\ \underline{ \bf \: \: \: multiplying \: \: both \: \: sides \: \: by \: \: \frac{ {c}^{2} }{ {(b - d)}^{2} } } \\ \\ = > \boxed{ \bf \frac{ {(a -c)}^{2} }{ {(b - d)}^{2} } = \frac{ {c}^{2} }{ {d}^{2} } } \: \: .........(2)$
Now...

Again..

$= > \frac{a}{c} = \frac{b}{d} \\ \\ \underline{ \: \: \: \bf adding \: \: 1 \: \: on \: \: both \: \: sides} \\ \\ = > \frac{a}{c} + 1 = \frac{b}{d} + 1 \\ \\ = > \frac{a + c}{c} = \frac{b + d}{d} \\ \\ \underline{\bf \: \: \: cubing \: \: both \: \: sides} \\ \\ = > {( \frac{a + c}{c} )}^{3} = {( \frac{b + d}{d} )}^{3} \\ \\ = > \frac{ {(a + c)}^{3} }{ {c}^{3} } = \frac{{(b + d)}^{3}}{ {d}^{3} } \\ \\ \bf \underline{ mutiplying \: \: both \: \: sides \: \: by \: \: \frac{ {c}^{3} }{ {(b + d)}^{3} } } \\ \\ = > \frac{ {(a + c)}^{3} }{ {(b + d)}^{3} } = \frac{ {c}^{3} }{ {d}^{3 } } \\ \\ = > \frac{ {(a + c)}^{3} }{ {(b + d)}^{3} } = \frac{c}{d} \times \frac{ {c}^{2} }{ {d}^{2} } \\ \\ \bf \underline{substituting \: \: the \: values \: \: from \: (1) \: \: and \: (2)} \\ \\ = > \frac{ {(a + c)}^{3} }{{(b + d) }^{3} } = \frac{a}{b} \times \frac{ {(a - c)}^{2} }{ {(b - d)}^{2} } \\ \\ = > \frac{ {(a + c)}^{3} }{ {(b + d)}^{3} } = \frac{a {(a - c)}^{2} }{b {(b - c)}^{2} } \\ \\ = > \boxed{ \bf {(a + c)}^{3} : { {(b + d)}^{3} } : \ \: : a {(a - c)}^{2} : b {(b - d)}^{2} }$
Hence.. Proved.
____________________________
Hope this is ur required answer

Proud to help you