Solve the following equation:
${{x}^{3}}-18x=35$ .

- Hint: Guess the root of the given cubic equation by hit and trial method and get a factor with help of the calculated root of the question. Calculate other factors by dividing the given cubic and roots by equating the other factor to ‘0’. Any quadratic $A{{x}^{2}}+Bx+c=0$ will have real root if $D={{B}^{{}}}-4AC\ge 0$ otherwise not.

Complete step-by-step solution -

Given equation in the problem is
$\begin{align}
  & {{x}^{3}}-18x=35 \\
 & \Rightarrow {{x}^{3}}-18x-35=0...................\left( i \right) \\
\end{align}$
As, there is not any particular method of getting roots of cubic polynomials. So, we need to find or guess one root by hit and trial method for factorizing the polynomial. So, let us put values of ‘x’ which may satisfy the given polynomial. Putting x = 2 to the equation (i), we get
$\begin{align}
  & {{\left( 2 \right)}^{3}}-18\times 2-35 \\
 & 8-36-35=-63\ne 0 \\
\end{align}$
Put x = 4 to the equation (i), we get
$\begin{align}
  & {{\left( 4 \right)}^{3}}-18\times 4-35 \\
 & 64-72-35=-8-35=-43\ne 0 \\
\end{align}$
Put x = 5 to the equation (i), we get
$\begin{align}
  & {{\left( 5 \right)}^{3}}-18\times 5-35 \\
 & 125-90-35=125-125=0 \\
\end{align}$
Hence, x = 5 is satisfying the given equation in the problem. It means x = 5 is the root of the polynomial and (x – 5) will be a factor of the given cubic equation.
 So, we can divide the given polynomial by (x – 5) to get another factor of the polynomial as the remainder by dividing the polynomial by (x – 5) will be 0. So, we get
$x-5\overset{{{x}^{2}}+5x+7}{\overline{\left){\begin{align}
  & \underline{\begin{align}
  & {{x}^{3}}-18x-35 \\
 & {{x}^{3}}-5{{x}^{2}} \\
\end{align}} \\
 & \underline{\begin{align}
  & 5{{x}^{2}}-18x-35 \\
 & 5{{x}^{2}}-25x \\
\end{align}} \\
 & 7x-35 \\
 & \underline{7x-35} \\
 & \text{ 0} \\
\end{align}}\right.}}$
Now, by the rule of division, we know that
$\text{dividend}=\text{divisor}\times \text{quotient}+\text{remainder}..................\left( ii \right)$
Hence we can write the given polynomial \[{{x}^{3}}-18x-35\] on dividing it by x – 5, we get
${{x}^{3}}-18x-35=\left( x-5 \right)\left( {{x}^{2}}+5x+7 \right)................(iii)$
Hence, we can write the equation (i) from equation (iii) as
$\left( x-5 \right)\left( {{x}^{2}}+5x+7 \right)=0$
As we know multiplication of two numbers can be 0, if any two of them is 0. So, we get
$\begin{align}
  & x-5=0\Rightarrow {{x}^{{}}}+5x+7=0 \\
 & x=5\Rightarrow {{x}^{2}}+5x+7=0 \\
\end{align}$
So, one root of the polynomial is x = 5 and other roots of the polynomial can be calculated by solving quadratic \[{{x}^{2}}+5x+7=0\] as mentioned above.
As, we know any quadratic $A{{x}^{2}}+bx+c=0$ will have real roots, if discriminant of the quadratic i.e. $D={{B}^{2}}-4AC$ is greater than 0 or equal to 0, but if D becomes less than 0, then roots will be unreal or imaginary. So, let us calculate discriminant of
${{x}^{2}}+5x+7=0\Rightarrow D={{B}^{2}}-4AC$
So, we get
$D={{\left( 5 \right)}^{2}}-4\times 1\times 7=25-28=-3$
Hence, we get D < 0. So, roots of the quadratic will not be real. Therefore, other two roots of the given cubic equation will not be real. So, real root of the given equation
${{x}^{2}}-18x-35=0\Rightarrow x=5$ .
Hence, x = 5 is the correct answer.

Note: One may follow another approach to get the other factor of ${{x}^{3}}-18x+35=0$ except x – 5 = 0 without using the division method. Here, we need to write (x – 5) to three times and try to arrange the terms in the following way:
\[{{x}^{2}}\left( x-5 \right)+5x\left( x-5 \right)-7\left( x-5 \right)={{x}^{3}}-18x+35\]
Here, we need to multiply (x – 5) by a term which on simplifying will give the polynomial. So, we can arrange the above expression as
$\left( x-5 \right)\left( {{x}^{2}}+5x-7 \right)={{x}^{3}}-18x+35$
One may go wrong if he or she takes (x + 5) as a factor of the given polynomial, which is the general mistake that can be made by the students. So, if 5 is a root then x – 5 should be a factor because on putting x – 5 to 0, we get value of x as 5, but on putting x + 5 to 0, we get x = -5, and x = -5 is not a root of the given polynomial. So, be clear with these concepts and don’t confuse yourself with the root and factor of any polynomial. One can calculate the other two imaginary roots with help of the quadratic ${{x}^{2}}+5x+7=0$, which is also a factor of the given polynomial. Roots of any quadratic $A{{x}^{2}}+bx+c=0$ , are given as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$
So, roots of given equation
$\begin{align}
  & x=\dfrac{-5\pm \sqrt{25-4\times 1\times 7}}{2} \\
 & x=\dfrac{-5\pm \sqrt{-3}}{2}=\dfrac{-5\pm \sqrt{3i}}{2} \\
 & i=\sqrt{-1} \\
\end{align}$