Question

If \[a + b + c = 1,{\text{ }}{a^2} + {b^2} + {c^2} = 2,{\text{ }}{a^3} + {b^3} + {c^3} = 3\] then find the value of \[{a^4} + {b^4} + {c^4} = ?\]

Answer 1

Hint: We know that \[{a^4} + {b^4} + {c^4} = {\left( {{a^2} + {b^2} + {c^2}} \right)^2} - 2\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right)\] .So, in order to solve this, we will first find out the value of \[\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right)\] . Now, \[\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right)\] can be written as \[\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) = {\left( {ab + bc + ca} \right)^2} - 2abc\left( {b + c + a} \right)\] .For this, we will first need to find out the value of \[ab + bc + ca\] by using the concept \[{\left( {a + b + c} \right)^2} - \left( {{a^2} + {b^2} + {c^2}} \right) = 2\left( {ab + bc + ca} \right)\] and then the value of \[abc\] using the formula \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ab + bc + ca} \right)} \right)\] . After that, just substitute the values required accordingly and we get the required result.

Complete step by step answer:
We have given
\[a + b + c = 1,{\text{ }}{a^2} + {b^2} + {c^2} = 2,{\text{ }}{a^3} + {b^3} + {c^3} = 3\]
And we have to find the value of \[{a^4} + {b^4} + {c^4}\]
We know that
\[{a^4} + {b^4} + {c^4} = {\left( {{a^2} + {b^2} + {c^2}} \right)^2} - 2\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right){\text{ }} - - - \left( 1 \right)\]
Now, let’s fist find out the value of \[\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right)\]
Now, \[\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right)\] can be written as
\[\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) = {\left( {ab + bc + ca} \right)^2} - 2abc\left( {b + c + a} \right){\text{ }} - - - \left( 2 \right)\]
We know that
\[{\left( {a + b + c} \right)^2} - \left( {{a^2} + {b^2} + {c^2}} \right) = 2\left( {ab + bc + ca} \right){\text{ }} - - - \left( 3 \right)\]
Now it is given that
\[a + b + c = 1\] and \[{a^2} + {b^2} + {c^2} = 2\]
So, on substituting the values in equation \[\left( 3 \right)\] we get
\[{\left( 1 \right)^2} - \left( 2 \right) = 2\left( {ab + bc + ca} \right)\]
\[ \Rightarrow 2\left( {ab + bc + ca} \right) = - 1\]
On dividing by \[2\] we get
\[ \Rightarrow \left( {ab + bc + ca} \right) = \dfrac{{ - 1}}{2}{\text{ }} - - - \left( X \right)\]
Now we know that
\[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ab + bc + ca} \right)} \right){\text{ }} - - - \left( 4 \right)\]
It is given that
\[a + b + c = 1,{\text{ }}{a^2} + {b^2} + {c^2} = 2\] and \[{a^3} + {b^3} + {c^3} = 3\]
So, on substituting the values in the equation \[\left( 4 \right)\] we get
\[3 - 3abc = \left( 1 \right)\left( {2 - \left( {ab + bc + ca} \right)} \right)\]
\[ \Rightarrow 3 - 3abc = \left( 1 \right)\left( {2 - \left( {\dfrac{{ - 1}}{2}} \right)} \right)\] from equation \[\left( X \right)\]
After simplification, we get
\[ \Rightarrow 3 - 3abc = \dfrac{5}{2}\]
\[ \Rightarrow 3 - \dfrac{5}{2} = 3abc\]
On subtracting the terms on the left-hand side, we get
\[ \Rightarrow \dfrac{1}{2} = 3abc\]
On dividing by \[3\] we get
\[abc = \dfrac{1}{6}{\text{ }} - - - \left( Y \right)\]
Now on substituting the values from equation \[\left( X \right)\] and equation \[\left( Y \right)\] in equation \[\left( 2 \right)\] we get
\[\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) = {\left( {\dfrac{{ - 1}}{2}} \right)^2} - 2\left( {\dfrac{1}{6}} \right)\left( 1 \right){\text{ }}\]
\[ \Rightarrow \left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) = \dfrac{1}{4} - \left( {\dfrac{1}{3}} \right)\]
On subtracting the terms on the right-hand side, we get
\[ \Rightarrow \left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) = \dfrac{{ - 1}}{{12}}{\text{ }} - - - \left( Z \right)\]
Now on substituting the value from equation \[\left( Z \right)\] in equation \[\left( 1 \right)\] we get
\[{a^4} + {b^4} + {c^4} = {\left( 2 \right)^2} - 2\left( {\dfrac{{ - 1}}{{12}}} \right){\text{ }}\]
\[ \Rightarrow {a^4} + {b^4} + {c^4} = 4 + \dfrac{1}{6}\]
On adding the terms on the right-hand side, we get
\[ \Rightarrow {a^4} + {b^4} + {c^4} = \dfrac{{25}}{6}\]
Hence, we get the required result.

Note:
 The first mistake students make while solving this question is to answer by looking at the given pattern in the question. i.e., in the question we have given \[a + b + c = 1,{\text{ }}{a^2} + {b^2} + {c^2} = 2,{\text{ }}{a^3} + {b^3} + {c^3} = 3\] .That’s why most students think the value of \[{a^4} + {b^4} + {c^4}\] will be \[4\] but this is the wrong approach to the question. Always try to solve the question using the common identities.