If x-a-y-b-z-c- then prove that x3-a3-y3-b3-z3-c3-3xyz-abc-

xa = yb = zc = l [(xa)^3 + (yb)^3 + (zc)^3 3 xyzabc] = (xa + yb + zc) [(xa)^2 + (yb)^2 + (zc)^2 xyab yzbc xzac] ⇒(xa)^3 + (yb)^3 + (zc)^3 ...

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Finding Square of a Number Using Identity

If x/a=y/b=...

Question

If $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$ , then prove that $\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3 x y z}{a b c}$ .

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\frac{x}{a} = \frac{y}{b} = \frac{z}{c}

\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3 x y z}{a b c}

\frac{x}{a} = \frac{y}{b} = \frac{z}{c}

\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3 {(x + y + z)}^{3}}{{(a + b + c)}^{3}}

(a^{3} + b^{3} + c^{3} - 3 a b c) (x^{3} + y^{3} + z^{3} - 3 x y z) = (X^{3} + Y^{3} + Z^{3} - 3 X Y Z)

x = a + b; y = b + c; z = c + a

\frac{x^{3} + y^{3} + z^{3} - 3 x y z}{a^{3} + b^{3} + c^{3} - 3 a b c}

\frac{x}{a} = \frac{y}{b} = \frac{z}{c}

\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}}

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