Question

Solve the following equations:
\[3{{x}^{3}}-8x{{y}^{2}}+{{y}^{3}}+21=0\] and \[{{x}^{2}}\left( y-x \right)=1\]
(a) 1, 2
(b) -1, 3
(c) \[\sqrt[3]{\dfrac{1}{2}},3\sqrt[3]{\dfrac{1}{2}}\]
(d) \[\sqrt{\dfrac{3}{2}},\dfrac{1}{2}\]

Answer 1

Hint: We solve this problem simply by substituting the value of \['y'\] we get from the second equation to first equation to solve for \['x'\] then for each value of \['x'\] we find the corresponding value of \['y'\] to get the solutions of given two equations.

Complete step-by-step solution:
We are given that the both equations are
\[\begin{align}
& \Rightarrow 3{{x}^{3}}-8x{{y}^{2}}+{{y}^{3}}+21=0.........equation(i) \\
& \Rightarrow {{x}^{2}}\left( y-x \right)=1.........equation(ii) \\
\end{align}\]
Now, let us take the second equation and solve for \['y'\] in terms of \['x'\] as follows
\[\begin{align}
  & \Rightarrow {{x}^{2}}\left( y-x \right)=1 \\
 & \Rightarrow {{x}^{2}}y-{{x}^{3}}=1 \\
 & \Rightarrow y=\dfrac{1+{{x}^{3}}}{{{x}^{2}}}..........equation(iii) \\
\end{align}\]
Now by substituting the value of \['y'\] in equation (i) we get
\[\Rightarrow 3{{x}^{3}}-8x{{\left( \dfrac{1+{{x}^{3}}}{{{x}^{2}}} \right)}^{2}}+{{\left( \dfrac{1+{{x}^{3}}}{{{x}^{2}}} \right)}^{3}}+21=0\]
Now, by expanding the each term we get
\[\begin{align}
& \Rightarrow 3{{x}^{3}}-8x\left( \dfrac{1+{{x}^{6}}+2{{x}^{3}}}{{{x}^{4}}} \right)+\left( \dfrac{1+{{x}^{9}}+3{{x}^{3}}+3{{x}^{6}}}{{{x}^{6}}} \right)+21=0 \\
& \Rightarrow 3{{x}^{3}}-\left( \dfrac{8+8{{x}^{6}}+16{{x}^{3}}}{{{x}^{3}}} \right)+\left( \dfrac{1+{{x}^{9}}+3{{x}^{3}}+3{{x}^{6}}}{{{x}^{6}}} \right)+21=0 \\
\end{align}\]
Now by doing the LCM and writing all the terms we get
\[\Rightarrow \dfrac{3{{x}^{9}}-8{{x}^{3}}-8{{x}^{9}}-16{{x}^{6}}+1+{{x}^{9}}+3{{x}^{3}}+3{{x}^{6}}+21{{x}^{6}}}{{{x}^{6}}}=0\]
Now, by adding all the terms of same power and cross multiplying the above equation we get
\[\Rightarrow -4{{x}^{9}}+8{{x}^{6}}-5{{x}^{3}}+1=0\]
Now, let us divide the terms in such a way that we can get something in common as follows
\[\Rightarrow \left( 4{{x}^{9}}-4{{x}^{6}} \right)+\left( -4{{x}^{6}}+4{{x}^{3}} \right)+\left( {{x}^{3}}-1 \right)=0\]
Now by taking the common terms out from each bracket we get
\[\begin{align}
  & \Rightarrow 4{{x}^{6}}\left( {{x}^{3}}-1 \right)-4{{x}^{3}}\left( {{x}^{3}}-1 \right)+1\left( {{x}^{3}}-1 \right)=0 \\
 & \Rightarrow \left( {{x}^{3}}-1 \right)\left( 4{{x}^{6}}-4{{x}^{3}}+1 \right)=0 \\
\end{align}\]
We know that if \[a\times b=0\] then either \['a'\] or \['b'\] is equal to zero
By using the above condition we get
\[\begin{align}
  & \Rightarrow {{x}^{3}}-1=0 \\
 & \Rightarrow x=1 \\
\end{align}\]
Now, let us take the other term that is
\[\begin{align}
  & \Rightarrow \left( 4{{x}^{6}}-4{{x}^{3}}+1 \right)=0 \\
 & \Rightarrow {{\left( 2{{x}^{3}}-1 \right)}^{2}}=0 \\
 & \Rightarrow x=\sqrt[3]{\dfrac{1}{2}} \\
\end{align}\]
Now, let us find the values of \['y'\] corresponding to each value of \['x'\]
Let us substitute \[x=1\] in equation (iii) we get
\[\Rightarrow y=\dfrac{1+1}{1}=2\]
Now, by substituting \[x=\sqrt[3]{\dfrac{1}{2}}\] in equation (iii) we get
\[\begin{align}
  & \Rightarrow y=\dfrac{1+\dfrac{1}{2}}{{{\left( \dfrac{1}{2} \right)}^{\dfrac{2}{3}}}} \\
 & \Rightarrow y=3\sqrt[3]{\dfrac{1}{2}} \\
\end{align}\]
Therefore the solutions of given two equations are \[\left( 1,2 \right)\] and \[\left( \sqrt[3]{\dfrac{1}{2}},3\sqrt[3]{\dfrac{1}{2}} \right)\]
So, option (a) and option (c) are correct answers.

Note: Students may do mistake in the solving of \['x'\] that is we have
\[\Rightarrow \left( {{x}^{3}}-1 \right)\left( 4{{x}^{6}}-4{{x}^{3}}+1 \right)=0\]
Here, in the first term we get
\[\begin{align}
  & \Rightarrow {{x}^{3}}-1=0 \\
 & \Rightarrow x=1 \\
\end{align}\]
Students may stop here since there is a option in which \[x=1\]
But, we need to solve for all remaining values because sometimes there may be multiple choice answers in which you may get incorrect for not solving for remaining values.
So we need to continue the remaining solution to get all the possible solutions.