Question

If `x/a = y/b = z/c`, prove that `[(a^2x^2 + b^2y^2 + c^2z^2)/(a^2x + b^3y +c^3z)]^3 = "xyz"/"abc"`

Answer 1

`x/a = y/b = z/c`
∴ x = ak, y = bk, z = c

L.H.S. = `[(a^2x^2 + b^2y^2 + c^2z^2)/(a^2x + b^3y +c^3z)]^3`

= `[(a^2.a^2k^2 + b^2.b^2l^2 + c^2.c^2k^2)/(a^3.a.k + b^3.bk + c^3.ck)]^3`

= `[(a^4k^2 + b^4k^2 + c^4k^2)/(a^4k + b^4k + c^4k)]^3`

= `[(k^2(a^4 + b^4 + c^4))/(k(a^4 + b^4 + c^4))]^3` = k³

R.H.S. = `"xyz"/"abc"`

= `"ak.bk.ck"/"abc"` = k³
∴ L.H.S. = R.H.S.