Question

How to find \[{m^2} + 4{n^2} = ?\] from \[{m^3} - 12m{n^2} = 40\] $4{n^3} - 3{m^2}n = 10$.

Answer 1

Hint: To solve this question, we will first multiply the second equation by 2. Then we have to square and add both of the equations. Then, we have to simplify the LHS of the resulting equation using the suitable algebraic identity. Finally, on taking the cube roots on both the sides, we will get the required value of the expression.

Complete step-by-step solution:
According to the above question, the given equations are
\[{m^3} - 12m{n^2} = 40\] (1)
$4{n^3} - 3{m^2}n = 10$ (2)
Squaring both sides of the first equation, we get
\[{\left( {{m^3} - 12m{n^2}} \right)^2} = {\left( {40} \right)^2}\]
Expanding the LHS using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ we get
\[{\left( {{m^3}} \right)^2} - 2\left( {{m^3}} \right)\left( {12m{n^2}} \right) + {\left( {12m{n^2}} \right)^2} = {\left( {40} \right)^2}\]
$ \Rightarrow {m^6} - 24{m^4}{n^2} + 144{m^2}{n^4} = 1600$ (3)
Multiplying the second equation by $2$ we get
$2\left( {4{n^3} - 3{m^2}n} \right) = 10 \times 2$
$ \Rightarrow 8{n^3} - 6{m^2}n = 20$
Squaring both sides, we get
${\left( {8{n^3} - 6{m^2}n} \right)^2} = {\left( {20} \right)^2}$
Expanding the LHS using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ we get
\[{\left( {8{n^3}} \right)^2} - 2\left( {8{n^3}} \right)\left( {6{m^2}n} \right) + {\left( {6{m^2}n} \right)^2} = {\left( {20} \right)^2}\]
$ \Rightarrow 64{n^6} - 96{m^2}{n^4} + 36{m^4}{n^2} = 400$ (4)
Adding (3) and (4) we get
\[{m^6} - 24{m^4}{n^2} + 144{m^2}{n^4} + 64{n^6} - 96{m^2}{n^4} + 36{m^4}{n^2} = 1600 + 400\]
$ \Rightarrow {m^6} + 12{m^4}{n^2} + 48{m^2}{n^4} + 64{n^6} = 2000$
Now, we can write ${m^6} = {\left( {{m^2}} \right)^3}$ and $64{n^6} = {\left( {4{n^2}} \right)^3}$ in the above equation to get
${\left( {{m^2}} \right)^3} + 12{m^4}{n^2} + 48{m^2}{n^4} + {\left( {4{n^2}} \right)^3} = 2000$
Multiplying and dividing the second term by $4{n^2}$ and the third term by ${\left( {4{n^2}} \right)^2}$ we get
\[{\left( {{m^2}} \right)^3} + \left( {4{n^2}} \right)\dfrac{{\left( {12{m^4}{n^2}} \right)}}{{\left( {4{n^2}} \right)}} + {\left( {4{n^2}} \right)^2}\dfrac{{\left( {48{m^2}{n^4}} \right)}}{{{{\left( {4{n^2}} \right)}^2}}} + {\left( {4{n^2}} \right)^3} = 2000\]
\[ \Rightarrow {\left( {{m^2}} \right)^3} + \left( {4{n^2}} \right)\left( {3{m^4}} \right) + {\left( {4{n^2}} \right)^2}\left( {3{m^2}} \right) + {\left( {4{n^2}} \right)^3} = 2000\]
Taking the factor of $3$ out from the second and the third terms we get
\[{\left( {{m^2}} \right)^3} + 3\left( {4{n^2}} \right)\left( {{m^4}} \right) + 3{\left( {4{n^2}} \right)^2}\left( {{m^2}} \right) + {\left( {4{n^2}} \right)^3} = 2000\]
Writing \[\left( {{m^4}} \right) = {\left( {{m^2}} \right)^2}\] we get
\[{\left( {{m^2}} \right)^3} + 3\left( {4{n^2}} \right){\left( {{m^2}} \right)^2} + 3{\left( {4{n^2}} \right)^2}\left( {{m^2}} \right) + {\left( {4{n^2}} \right)^3} = 2000\]
From the identity ${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$ we can write the LHs of the above equation as
\[{\left( {{m^2} + 4{n^2}} \right)^3} = 2000\]
Taking cube root both the sides, we get
\[{m^2} + 4{n^2} = 10\sqrt[3]{2}\]

Hence, the final answer is \[10\sqrt[3]{2}\].

Note:
For solving such types of questions, we must be familiar with the algebraic identities such as ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and ${\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$. From the given equations, we cannot explicitly evaluate the values of $m$ and $n$. Therefore, we can only use algebraic manipulations for solving such questions. An algebraic equation is an equation which is a combination of variables and its coefficients and also includes constant terms and equal to sign. If the equal to sign is not there then the equation changes to algebraic expression.