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Mathematics

If x = a+13+a−13a+13−a−13,\dfrac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} - \sqrt[3]{a - 1}},

prove that x3 - 3ax2 + 3x - a = 0.

Ratio Proportion

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Answer

Given,

x1=a+13+a−13a+13−a−13\dfrac{x}{1} = \dfrac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} - \sqrt[3]{a - 1}}

Applying componendo and dividendo,

⇒x+1x−1=a+13+a−13+a+13−a−13a+13+a−13−a+13+a−13⇒x+1x−1=2a+132a−13\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{a + 1} + \sqrt[3]{a - 1} + \sqrt[3]{a + 1} - \sqrt[3]{a - 1}}{\sqrt[3]{a + 1} + \sqrt[3]{a - 1} - \sqrt[3]{a + 1} + \sqrt[3]{a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt[3]{a + 1}}{2\sqrt[3]{a - 1}}

Cubing both the sides,

⇒(x+1)3(x−1)3=a+1a−1\Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} = \dfrac{a + 1}{a - 1}

Again applying componendo and dividendo,

⇒(x+1)3+(x−1)3(x+1)3−(x−1)3=a+1+a−1a+1−a+1⇒x3+1+3x(x+1)+x3−1−3x(x−1)x3+1+3x(x+1)−(x3−1−3x(x−1))=2a2⇒x3+x3+1−1+3x2−3x2+3x+3xx3−x3+1+1+3x2+3x2+3x−3x=2a2⇒2x3+6x2+6x2=a⇒2(x3+3x)2(1+3x2)=a⇒x3+3x1+3x2=a\Rightarrow \dfrac{(x + 1)^3 + (x - 1)^3}{(x + 1)^3 - (x - 1)^3} = \dfrac{a + \cancel{1} + a - \cancel{1}}{\cancel{a} + 1 - \cancel{a} + 1} \\[1em] \Rightarrow \dfrac{x^3 + 1 + 3x(x + 1) + x^3 - 1 - 3x(x - 1)}{x^3 + 1 + 3x(x + 1) - (x^3 - 1 - 3x(x - 1))} = \dfrac{2a}{2} \\[1em] \Rightarrow \dfrac{x^3 + x^3 + 1 - 1 + 3x^2 - 3x^2 + 3x + 3x}{x^3 - x^3 + 1 + 1 + 3x^2 + 3x^2 + 3x - 3x} = \dfrac{2a}{2} \\[1em] \Rightarrow \dfrac{2x^3 + 6x}{2 + 6x^2} = a \\[1em] \Rightarrow \dfrac{2(x^3 + 3x)}{2(1 + 3x^2)} = a \\[1em] \Rightarrow \dfrac{x^3 + 3x}{1 + 3x^2} = a

On cross-multiplication,

⇒x3+3x=a(1+3x2)⇒x3+3x=a+3ax2⇒x3−3ax2+3x−a=0.\Rightarrow x^3 + 3x = a(1 + 3x^2) \\[1em] \Rightarrow x^3 + 3x = a + 3ax^2 \\[1em] \Rightarrow x^3 - 3ax^2 + 3x - a = 0.

Hence, proved that x3 - 3ax2 + 3x - a = 0.

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