Shew that $a^{3} (b z - c y)^{3} + b^{3} (c x - a z)^{3} + c^{3} (a y - b x)^{3} = 3 a b c (b z - c y) (c x - a z) (a y - b x)$ .

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Solution

Assuming $a (b z - c y) = A, b (c x - a z) = B, c (a y - b b x) = C$ , We want to prove that

$A^{3} + B^{3} + C^{3} = 3 A B C$

Now, $A + B + C = 0$ and $A^{3} + B^{3} + C^{3} - 3 A B C = (A + B + C) (A^{2} + B^{2} + C^{2} - A B - B C - C A) = 0$

$∴ A^{3} + B^{3} + C^{3} = 3 A B C$

Hence proved

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∣ ∣ ∣ ∣ ∣ \begin{matrix} l^{2} - m^{2} & a - b & x^{3} - y^{3} m^{2} - n^{2} & b - c & y^{3} - z^{3} n^{2} - l^{2} & c - a & z^{3} - x^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a x - b y - c z & a y + b x & c x + a z a y + b x & b y - c z - a x & b z + c y c x + a z & b z + c y & c z - a x - b y \end{matrix} ∣ ∣ ∣ ∣

= (x^{2} + y^{2} + z^{2}) (a^{2} {+ b}^{2} {+ c}^{2}) (a x + b y + c z) .

A X_{3}, B X_{3}

C Y_{3}

4.97 \times 10^{- 30}, 0.60 \times 10^{- 30}

0.00

\frac{x^{2}}{1 - a^{2}} = \frac{y^{2}}{1 - b^{2}} = \frac{z^{2}}{1 - c^{2}} .

x^{4} - (a + b) x^{3} - a (a - b) x^{2} + a^{2} (a + b) x - a^{3} b = 0

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