Question

(a^2+ac +c^2):(a^2-ac+c^2)=(b^2+bd+d^2):(b^2-bd+d^2) prove that if a:b::c:d​

Answer 1

Answer:

Did you mean $a:b=c:d$?

Step-by-step explanation:

Identity

( A ± B ){ A^2 - (±AB) + B^2 } = A^3 ± B^3

$(a^2+ac +c^2):(a^2-ac+c^2)=a^3-c^3:a^3+c^3$

$(b^2+bd+d^2):(b^2-bd+d^2)=b^3-d^3:b^3+b^3$

$a^3-c^3:a^3+c^3=b^3-d^3:b^3+d^3\\(a^3+c^3)(b^3-d^3)=(a^3-c^3)(b^3+d^3)$

$a^3b^3-a^3d^3+b^3c^3-c^3d^3=a^3b^3+a^3d^3-b^3c^3-c^3d^3\\(a^3b^3-a^3b^3)+(b^3c^3+b^3c^3)=(a^3d^3+a^3d^3)-(c^3d^3+c^3d^3)\\2a^3d^3=2b^3c^3\\a^3d^3=b^3c^3\\ad=bc$

∴$ad=bc$

In $a:b=c:d$,

$ad=bc$

∴$a:b=c:d$