QUESTION

# Solve the following equations:$\left( {{x^2} - {y^2}} \right)\left( {x - y} \right) = 16xy,\left( {{x^4} - {y^4}} \right)\left( {{x^2} - {y^2}} \right) = 640{x^2}{y^2}$ ${\text{A}}{\text{. }}x{\text{ = 0,}}y{\text{ = 0}} \\ {\text{B}}{\text{. }}x{\text{ = 9,}}y{\text{ = 3}} \\ {\text{C}}{\text{. }}x{\text{ = 4,}}y{\text{ = 5}} \\ {\text{D}}{\text{. }}x{\text{ = 3,}}y{\text{ = 9}} \\$

Hint: If you depict these equations as quadratic i.e. equation with degree $0$ and biquadratic i.e. equation with degree $1$ in two variables, you are absolutely on the right path. You must write both the equations and try to find the common term between them to make them connected in a certain way as in order to solve them which will then lead us to the values of $x{\text{ and }}y$.

Given equations are:
$\left( {{x^2} - {y^2}} \right)\left( {x - y} \right) = 16xy$
$\Rightarrow xy = \dfrac{{\left( {{x^2} - {y^2}} \right)\left( {x - y} \right)}}{{16}}$ … (1)
And squaring on both sides we get,
$\left( {{x^4} - {y^4}} \right)\left( {{x^2} - {y^2}} \right) = 640{x^2}{y^2}$
By using formula ${a^4} - {b^4} = \left( {a - b} \right)\left( {a + b} \right)\left( {{a^2} + {b^2}} \right)$ we get,
$\Rightarrow \left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right) = 640{x^2}{y^2}$
Now by using formula $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ we can write,
$\Rightarrow \left( {{x^2} + {y^2}} \right){\left( {{x^2} - {y^2}} \right)^2} = 640{\left( {xy} \right)^2}$
Substitute the value of equation (1), in above equation, we get
$\Rightarrow \left( {{x^2} + {y^2}} \right){\left( {{x^2} - {y^2}} \right)^2} = 640{\left\{ {\dfrac{{\left( {{x^2} - {y^2}} \right)\left( {x - y} \right)}}{{16}}} \right\}^2}$
$\Rightarrow \left( {{x^2} + {y^2}} \right){\left( {{x^2} - {y^2}} \right)^2} = 640\left\{ {\dfrac{{{{\left( {{x^2} - {y^2}} \right)}^2}{{\left( {x - y} \right)}^2}}}{{256}}} \right\}$
$\Rightarrow \,2\left( {{x^2} + {y^2}} \right){\left( {{x^2} - {y^2}} \right)^2} = 5{\left( {{x^2} - {y^2}} \right)^2}{\left( {x - y} \right)^2} \\ \Rightarrow \,2\left( {{x^2} + {y^2}} \right) = 5{\left( {x - y} \right)^2} \\$
Now by using formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$,
$\Rightarrow \,2\left( {{x^2} + {y^2}} \right) = 5\left( {{x^2} + {y^2} - 2xy} \right) \\ \Rightarrow {\text{2}}{x^2} + 2{y^2} = 5{x^2} + 5{y^2} - 10xy \\ \Rightarrow 3{x^2} + 3{y^2} = 10xy \\ \Rightarrow {x^2} + {y^2} = \dfrac{{10xy}}{3} \\$
Subtract $2xy$ from both sides,
$\Rightarrow {x^2} + {y^2} - 2xy = \dfrac{{10xy}}{3} - 2xy$
Now by using ${a^2} - {b^2} + 2ab = {\left( {a - b} \right)^2}$,
$\Rightarrow {\left( {x - y} \right)^2} = \dfrac{{4xy}}{3}{\text{ }}$ … (2)
By using formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ from (1) we have,
$\Rightarrow \left( {x + y} \right){\left( {x - y} \right)^2} = 16xy$ … (3)
Substituting (2) in (3) we get,
$\left( {x + y} \right)\dfrac{{4xy}}{3} = 16xy \\ \Rightarrow \;xy\left( {x + y} \right) = 12xy \\ \Rightarrow xy\left( {x + y} \right) - 12xy = 0 \\ \Rightarrow xy\left( {x + y - 12} \right) = 0 \\ {\text{If }}xy = 0 \\ \Rightarrow x = 0,y = 0 \\ {\text{Also }}x + y - 12 = 0 \\ \Rightarrow y = 12 - x \\$
Substituting $y$ in (3) we get,
$\left( {x + 12 - x} \right){\left( {x - 12 + x} \right)^2} = 16x\left( {12 - x} \right) \\ \Rightarrow 12{\left( {2x - 12} \right)^2} = 16\left( {12x - {x^2}} \right) \\ \Rightarrow {\text{3}} \times 4{\left( {x - 6} \right)^2} = 4\left( {12x - {x^2}} \right) \\ \Rightarrow 3{\left( {x - 6} \right)^2} = 12x - {x^2} \\$
Now use ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ formula,
$\Rightarrow 3{x^2} + 108 - 36x = 12x - {x^2} \\ \Rightarrow 4{x^2} - 48x + 108 = 0 \\$
By dividing the whole equation by $4$ we get,
${x^2} - 12x + 27 = 0$
Now we will use factorisation method to solve this equation,
${x^2} - 9x - 3x + 27 = 0 \\ \Rightarrow x\left( {x - 9} \right) - 3\left( {x - 9} \right) = 0 \\ \Rightarrow \left( {x - 3} \right)\left( {x - 9} \right) = 0 \\ \Rightarrow x = 3,9 \\ \therefore {\text{ }}y = 12 - x \\ \Rightarrow y = 9,3{\text{ }} \\$
So, the values of $x$ are $0,3,9$ and corresponding values of $y$ are $0,9,3$
The combinations are:
$\left( {0,0} \right) \\ \left( {3,9} \right) \\ \left( {9,3} \right) \\$

Note: Whenever you have to solve these types of questions, firstly solve one of the two equations for one of the variables in terms of the other and then substitute the expression for this variable into the second equation. Then solve for the remaining variable, then we simply substitute that solution into either of the original equations to find the value of the first variable. Thus, we get our desirable answer i.e. the values of $x$ and $y$.