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Question
If `x/a = y/b = z/c`, prove that `x^3/a^2 + y^3/b^2 + z^3/c^2 = (x+ y+ z)^3/(a + b+ c)^2`
Solution
`x/a = y/b = z/c`
∴ x = ak, y = bk, z = ck
L.H.S. = `x^3/a^2 + y^3/b^2 + z^3/c^2`
= `(a^3k^3)/a^2 + (b^3k^3)/(b^2) + (c^3k^3)/c^2`
= ak3 + bk3 + ck3
= k3(a + b + c)
R.H.S. = `(x + y + z)^3/(a + b + c)^2`
= `(ak + bk + ck)^3/(a + b + c)^2`
= `(k^3(a + b + c)^3)/((a + b + c)`
= k3(a + b + c)
Hence L.H.S. = R.H.S.
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