Question

Solve the equation and obtain the value of x from the following equation
${{x}^{3}}+3{{x}^{2}}+3x-7=0$

Answer 1

Hint: The given equation is a cubic equation of x. Therefore, we should try to factorize the Left Hand Side of the equation into a linear and a quadratic factor. As the RHS is zero and the LHS is zero whenever any one of its factors is zero, we can equate the factors to zero to find the solutions to the given equation.

Complete step-by-step answer:
The given equation is
${{x}^{3}}+3{{x}^{2}}+3x-7=0$
To factorize the LHS, we note that we should try to find the factors of the constant term and divide them by the coefficient of the cubic term and then look if any of these numbers or their negatives satisfies the equation, i.e. the value of the LHS becomes zero if we take the value of x to be that number.
We know that the factors of 7 are 1 and 7 and their negatives are -1 and -7 respectively. Now, if we take x= 1, 7, -1, 7 we obtain the LHS as
${{1}^{3}}+3\times {{1}^{2}}+3\times 1-7=0$
Therefore, as x=1 satisfies the given equation, (x-1) should be a factor of the LHS. Therefore, we can divide ${{x}^{3}}+3{{x}^{2}}+3x-7$ by (x-1) to obtain the other factor.
$x-1\overset{{{x}^{2}}-2x+1}{\overline{\left){\begin{align}
  & {{x}^{3}}+3{{x}^{2}}+3x-1 \\
 & {{x}^{3}}-{{x}^{2}} \\
 & 0+2{{x}^{2}}+3x-1 \\
 & \text{ }2{{x}^{2}}+2x \\
 & \text{ }x-1\text{ } \\
 & \text{ }x-1 \\
 & \text{ }0 \\
\end{align}}\right.}}$
Thus, we can write the expression in LHS as ${{x}^{3}}+3{{x}^{2}}+3x-7=(x-1)({{x}^{2}}-2x+1)$. Thus, the expression will be zero if $x-1=0\Rightarrow x=1...............(1.1)$ or when ${{x}^{2}}-2x+1=0............(1.2)$
We know that the solution of a general quadratic equation $a{{x}^{2}}+bx+c=0$ is
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Thus, using this formula with a=1, b=-2 and c=1, the solution of ${{x}^{2}}-2x+1=0$ will be zero when
$x=\dfrac{2\pm \sqrt{-{{2}^{2}}-4\times 1\times 1}}{2\times 1}=1.........(1.3)$
Thus, from equations of (1.1), (1.2) and (1.3), we find the solution of the given equation as x=1,1,1. Thus, the equation is satisfied only for x=1.

Note: We should note that even though the equation was cubic, the solution was only the point x=1. This is not a problem because in a cubic equation, there should be 3 roots but they may occur at the same point. Also, using the formula ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$ with a=x and b=-1, we can write the given equation as ${{x}^{3}}+3{{x}^{2}}+3x-7={{\left( x-1 \right)}^{3}}=0$ and thus, it has the solution at $x-1=0\Rightarrow x=1$ as obtained in the solution.