Question

If `x/a = y/b = z/c`, prove that: `(x^2a^2 + y^2b^2 + z^2c^2)/(a^3x + b^3y + c^3z) = ((xyz)/(abc))^(1/3)`

Answer 1

`x/a = y/b = z/c` = k

x = ak, y = bk, z = ck

L.H.S.

= `(x^2a^2 + y^2b^2 + z^2c^2)/(a^3x + b^3y + c^3z)`

= `((ak)^2a^2 + (bk)^2b^2 + (ck)^2c^2)/(a^3(ak) + b^3(bk) + c^3(ck))`

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

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

= `k^2/k`

= k

R.H.S.

= `((xyz)/(abc))^(1/3)`

= `(((ak)(bk)(ck))/(abc))^(1/3)`

= `((abck^3)/(abc))^(1/3)`

= `(k^3)^(1/3)`

= k

L.H.S. = R.H.S.