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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)`.
Solution
Let `x/a=y/b = z/c` = k
=> 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`
`= k^3 + k^3 + k^3`
= `3k^3`
R.H.S = `(3xyz)/(abc)`
`= (3(ak)(bk)(ck))/(abc)`
`= (3(k^3)(abc))/(abc)`
`= 3k^3`
= L.H.S
=> L.H.S = R.H.S
`=> x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`
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