Question

If `x/a = y/b = z/c` show that: `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.

Answer 1

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)`