If `a : b = c : d = e : f`, then prove that ` (a^(2) + c^(2) + e^(2)) (b^(2)+d^(2)+f^(2)) = (ab + cd + ef)^(2))`

Explanation

Given that ` a : b = c : d = e : f`
or, `a/b = c/d = e/f = k " (let)" ( k ne 0)`
` :. A = bk , c = dk " and " e = fk`.
Now, LHS ` = (a^(2) + c^(2) + e^(2)) (b^(2) + d^(2) + f^(2))`
` = {(bk)^(2) + (dk)^(2) + (fk)^(2)} (b^(2) + d^(2) + f^(2))`
` = (b^(2)k^(2) + d^(2) k^(2) + f^(2) k^(2))(b^(2) + d^(2) + f^(2))`
` = k^(2) ( b^(2) + d^(2) + f^(2)) (b^(2) + d^(2) + f^(2))`
` = k^(2) (b^(2)+d^(2)+f^(2))^(2)`
RHS ` = (ab+cd+ef)^(2) = (bk.b + dk.d + fk.f)^(2)`
` = (b^(2)k+d^(2)k+f^(2)k)^(2) = {k(b^(2)+d^(2) +f^(2))}^(2)`
` = k^(2) (b^(2) +d^(2)+f^(2))^(2)`.
` :. ` LHS = RHS