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Question
If `x/a = y/b = z/c` show that: `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.
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Solution
Let `x/a = y/b = z/c = k`
`\implies` x = ak, y = bk, z = ck
L.H.S = `x^3/a^3 + y^3/b^3 + z^3/c^3`
= `(ak)^3/a^3 + (bk)^3/b^3 + (ck)^3/c^3`
= `(a^3k^3)/a^3 + (b^3k^3)/b^3 + (c^3k^3)/c^3`
= k3 + k3 + k3
= 3k3
R.H.S = `(3xyz)/(abc)`
= `(3(ak)(bk)(ck))/(abc)`
= 3k3
`\implies` L.H.S = R.H.S
i.e `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`
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