Question

If a and b are the solutions of the equation \[{x^2} + mx - m = 0\], find the value of \[{a^3} + {b^3}\] in terms of m.

Answer 1

Hint: We use the formula of roots of a quadratic equation and find the values of ‘a’ and ‘b’. From the formula of \[{(a + b)^3}\], calculate the formula for \[{a^3} + {b^3}\]and substitute the values of ‘a’ and ‘b’ in it. Solve using arithmetic operations. Use the property \[(x + y)(x - y) = {x^2} - {y^2}\] to solve for the solution.
* \[{(x + y)^3} = {x^3} + {y^3} + 3xy(x + y)\]
* The roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step solution:
We are given the quadratic equation \[{x^2} + mx - m = 0\]..............… (1)
We know the roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Compare the equation given i.e. \[{x^2} + mx - m = 0\] with the general quadratic equation we get
\[a = 1,b = m,c = - m\]
\[ \Rightarrow \]Roots of equation \[{x^2} + mx - m = 0\] are \[x = \dfrac{{ - m \pm \sqrt {{m^2} - 4 \times 1 \times ( - m)} }}{{2 \times 1}}\]
Solve the value under the square root
\[ \Rightarrow \]Roots of equation \[{x^2} + mx - m = 0\] are \[x = \dfrac{{ - m \pm \sqrt {{m^2} + 4m} }}{2}\]
\[ \Rightarrow \]Roots of equation \[{x^2} + mx - m = 0\] are \[\dfrac{{ - m + \sqrt {{m^2} + 4m} }}{2}\] and \[\dfrac{{ - m - \sqrt {{m^2} + 4m} }}{2}\]..............… (2)
Since we are given that ‘a’ an ‘b’ are roots of the equation \[{x^2} + mx - m = 0\]
We name one of the roots obtained in equation (2) as ‘a’ and one root as ‘b’
\[a = \dfrac{{ - m + \sqrt {{m^2} + 4m} }}{2}\]and \[b = \dfrac{{ - m - \sqrt {{m^2} + 4m} }}{2}\] …………..… (3)
Now we know the identity \[{(x + y)^3} = {x^3} + {y^3} + 3xy(x + y)\].
i.e. \[{(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)\]
Shift the values except \[{a^3} + {b^3}\] on one side of the equation
\[ \Rightarrow {(a + b)^3} - 3ab(a + b) = {a^3} + {b^3}\]
Substitute the values of ‘a’ and ‘b’ from equation (2) in the formula
\[ \Rightarrow {\left( {\dfrac{{ - m + \sqrt {{m^2} + 4m} }}{2} + \dfrac{{ - m - \sqrt {{m^2} + 4m} }}{2}} \right)^3} - 3\left( {\dfrac{{ - m + \sqrt {{m^2} + 4m} }}{2} \times \dfrac{{ - m - \sqrt {{m^2} + 4m} }}{2}} \right)\left( {\dfrac{{ - m + \sqrt {{m^2} + 4m} }}{2} + \dfrac{{ - m - \sqrt {{m^2} + 4m} }}{2}} \right) = {a^3} + {b^3}\]
Take LCM inside the brackets and solve for the value of addition. Also, use identity \[(x + y)(x - y) = {x^2} - {y^2}\] to solve multiplication bracket
\[ \Rightarrow {\left( {\dfrac{{ - m + \sqrt {{m^2} + 4m} - m - \sqrt {{m^2} + 4m} }}{2}} \right)^3} - \dfrac{3}{4}\left( {{{( - m)}^2} - {{\left( {\sqrt {{m^2} + 4m} } \right)}^2}} \right)\left( {\dfrac{{ - m + \sqrt {{m^2} + 4m} - m - \sqrt {{m^2} + 4m} }}{2}} \right) = {a^3} + {b^3}\]Cancel the possible terms in numerator
\[ \Rightarrow {\left( {\dfrac{{ - 2m}}{2}} \right)^3} - \dfrac{3}{4}\left( {{m^2} - \left( {{m^2} + 4m} \right)} \right)\left( {\dfrac{{ - 2m}}{2}} \right) = {a^3} + {b^3}\]
Cancel the same factors from numerator and denominator
\[ \Rightarrow - {m^3} - \dfrac{3}{4}\left( {{m^2} - {m^2} - 4m} \right)\left( { - m} \right) = {a^3} + {b^3}\]
Cancel the terms with same magnitude but opposite signs
\[ \Rightarrow - {m^3} - \dfrac{3}{4}\left( { - 4m} \right)\left( { - m} \right) = {a^3} + {b^3}\]
Cancel the same factors from numerator and denominator
\[ \Rightarrow - {m^3} - 3{m^2} = {a^3} + {b^3}\]

\[\therefore \]The value of \[{a^3} + {b^3}\] is \[ - {m^3} - 3{m^2}\]

Note: Many students make mistake of calculating the value of ‘a’ and ‘b’ using the different properties and then substitute in \[{a^3} + {b^3}\] to calculate its value which is very confusing and long process. Keep in mind we use the identity of the whole cube as we know these terms come directly in RHS of the identity.