Discussion Forum

The total number of distincts   $x \epsilon R$ for  which $\begin{bmatrix}x & x^{2} &1+x^{3} \\2x & 4x^{2}& 1+8x^{3} \\3x&9x^{2}&1+27x^{3} \end{bmatrix}=10$  is

Answer:

Explanation:

Given,    $\begin{bmatrix}x & x^{2} &1+x^{3} \\2x & 4x^{2}& 1+8x^{3} \\3x&9x^{2}&1+27x^{3} \end{bmatrix}=10$

  =  $\Rightarrow x.x^{2}\begin{bmatrix}1 & 1 &1+x^{3} \\2 & 4& 1+8x^{3} \\3&9&1+27x^{3} \end{bmatrix}=10$

 Apply  $R_{2}\rightarrow R_{2}-2R_{1}$  and $R_{3}\rightarrow R_{3}-3R_{1}$

we get,

              $x^{3}\begin{bmatrix}1 & 1 &1+x^{3} \\0 & 2& -1+6x^{3} \\0&6&-2+24x^{3} \end{bmatrix}=10$

  $\Rightarrow$     $x^{3}\begin{bmatrix}2 & 6x^{3} &-1\\6 & 24x^{3}&-2 \end{bmatrix}=10$

   $\Rightarrow$    $x^{3} (48x^{3}-4-36x^{3}+6)=10$

  $\Rightarrow$   $12x^{6}+2x^{3}=10$

   $\Rightarrow$    $6x^{6}+x^{3}-5=0$

  $\Rightarrow$  $x^{3}=\frac{5}{6},-1$

$\therefore$               $x=\left(\frac{5}{6}\right)^{1/3},-1$

   Hence, the number of real solutions is 2