Respuesta :
Solution:-
[tex] \text{Given Equation is } \frac{x}{1} = \frac{ \sqrt[3]{m + 1{ }} + \sqrt[3]{m - 1} }{ \sqrt[3]{m + 1} + \sqrt[3]{m - 1} }[/tex]
[tex]\text{On Applying componendo and dividendo, we get:-}[/tex]
[tex] \frac{x + 1}{x - 1} = \frac{( \sqrt[3]{m + 1} + \sqrt[3]{m - 1} )+ ( \sqrt[3]{m + 1}) - \sqrt[3]{m - 1} }{ (\sqrt[3]{m + 1} + \sqrt[3]{m - 1}) - ( \sqrt[3]{m + 1}) - \sqrt[3]{m - 1} } [/tex]
[tex] \implies \: \frac{x + 1}{x - 1} = \frac{2( \sqrt[3]{m + 1)} }{2( \sqrt[3]{m - 1)} } \\ \\ \implies \frac{ x + 1}{x - 1} = \frac{ \sqrt[3]{m + 1} }{ \sqrt[3]{m - 1} } [/tex]
[tex] \text{On Cubing both sides, we get} \\ \frac{(x + 1) {}^{3} }{(x - 1) {}^{3} } = \frac{m + 1}{m - 1} [/tex]
[tex]\text{Again, Applying componendo and dividendo, we get:-}[/tex]
[tex] \frac{(x + 1) {}^{3} + (x - 1) {}^{3} }{(x + 1) {}^{3} - (x - 1) {}^{3} } = \frac{m + 1 + m - 1}{m + 1 - (m - 1)} [/tex]
[tex] \implies \frac{ {x}^{3} + 1 {}^{3} + 3 {x}^{2} \times 1 + 3x \times {1}^{2} + ( {x}^{3} - 1 {}^{3} - 3x {}^{2} \times 1 + 3x \times 1 {}^{2} ) }{x {}^{3} + 1 {}^{3} + 3 {x}^{2} \times 1 + 3x \times {1}^{2} - (x {}^{3} - 1 {}^{3} - 3 {x}^{2} \times 1 + 3x \times {1}^{2}) } = \frac{2m}{2} [/tex]
[tex] \implies \frac{2x {}^{3} + 6x}{ {6x}^{2} + 2 } = \frac{m}{1} [/tex]
[tex] \implies \frac{2(x {}^{3 } + 3x) }{2(3 {x}^{2} + 1) } = \frac{m}{1} [/tex]
[tex] \implies \: x {}^{3} + 3x = 3m {x}^{2} + m \\ \\ \implies {x}^{3} - 3mx {}^{2} + 3x \\ = m[/tex]
Hope This Helps!!
Based on the calculations, the value of [tex]x^3-3mx^2+3x[/tex] is equal to m.
Given the following data:
[tex]x=\frac{\sqrt[3]{m+1} +\sqrt[3]{m-1}}{\sqrt[3]{m+1} +\sqrt[3]{m-1}}[/tex]
How to evaluate the fraction.
In order to simplify the given mathematical expression, we would apply the componendo and dividendo rule because it would the number of expansions as follows:
[tex]\frac{x}{1} =\frac{\sqrt[3]{m+1} +\sqrt[3]{m-1}}{\sqrt[3]{m+1} -\sqrt[3]{m-1}}\\\\\frac{x+1}{x-1} =\frac{(\sqrt[3]{m+1} +\sqrt[3]{m-1})+(\sqrt[3]{m+1} -\sqrt[3]{m-1})}{(\sqrt[3]{m+1} +\sqrt[3]{m-1})-(\sqrt[3]{m+1} -\sqrt[3]{m-1})}\\\\\frac{x+1}{x-1} =\frac{\sqrt[3]{m+1} +\sqrt[3]{m-1}+\sqrt[3]{m+1} -\sqrt[3]{m-1}}{\sqrt[3]{m+1} +\sqrt[3]{m-1}-\sqrt[3]{m+1} +\sqrt[3]{m-1}}\\\\\frac{x+1}{x-1} =\frac{2(\sqrt[3]{m+1})}{2(\sqrt[3]{m-1})} \\\\\frac{x+1}{x-1} =\frac{\sqrt[3]{m+1}}{\sqrt[3]{m-1}}[/tex]
Taking the cube of both sides, we have:
[tex](\frac{x+1}{x-1})^3 =(\frac{\sqrt[3]{m+1}}{\sqrt[3]{m-1}})^3\\\\\frac{(x+1)^3}{(x-1)^3}=\frac{m+1}{m-1}[/tex]
Applying the componendo and dividendo rule, we have:
[tex]\frac{(x+1)^3+(x-1)^3}{(x-1)^3-(x-1)^3}=\frac{m+1+(m-1)}{m-1-(m-1)}\\\\\frac{(x+1)^3+(x-1)^3}{(x-1)^3-(x-1)^3}=\frac{2m}{2} \\\\\frac{(x+1)^3+(x-1)^3}{(x-1)^3-(x-1)^3}=m\\\\\frac{2x^3+6x}{6x^2+2} =m\\\\\frac{2(x^3+3x)}{2(3x^2+1)} =m\\\\\frac{x^3+3x}{3x^2+1} =m\\\\x^3+3x=m(3x^2+1)\\\\x^3+3x=3mx^2+m\\\\x^3-3mx^2+3x=m[/tex]
Read more on componendo and dividendo here: https://brainly.com/question/134857
