Question

If $ x+y+z=12 $ and $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=96 $ and $ \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=36 $ then the value of $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ is?

Answer 1

Hint: Here, we will use the method of factorization to factorize $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $. This will help us to write $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ as a product of two factors comprising of the expressions whose value is given to us in the question. We will use algebraic identities like $ {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac $, $ {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}} $ and $ {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{3}}+{{b}^{3}}-ab \right) $ in our calculations.

Complete step-by-step answer:
Since, it is given that:
 $ x+y+z=12 $
On squaring both sides and using the identity $ {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac $ , we get:
\[\begin{align}
  & {{\left( x+y+z \right)}^{2}}={{\left( 12 \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx=144 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( xy+yz+zx \right)=144............\left( 1 \right) \\
\end{align}\]
Also, it is given that:
 $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=96...........\left( 2 \right) $
And, also $ \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=36 $
On taking the LCM of x, y and z and simplifying it, we get:
 $ \dfrac{yz+xz+xy}{xyz}=36 $
 $ \Rightarrow xy+yz+zx=36xyz.........\left( 3 \right) $
On putting values from equation (2) and equation (3) in equation (1), we get:
 $ \begin{align}
  & 96+2\left( 36xyz \right)=144 \\
 & \Rightarrow 72xyz=144-96 \\
 & \Rightarrow 72xyz=48 \\
 & \Rightarrow xyz=\dfrac{48}{72}=\dfrac{2}{3}............\left( 4 \right) \\
\end{align} $
Now, we can write $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ as:
 $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}-3x{{y}^{2}}-3{{x}^{2}}y-3xyz+3xyz $
(Here, we have added and subtracted same expressions, so the value of $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ will remain same)
On taking (-3xy) common from $ \left( -3{{x}^{2}}y,-3x{{y}^{2}}\text{ and }-3xyz \right) $ , we get:
$ {{x}^{3}}+{{y}^{3}}+{{z}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}-3xy\left( x+y+z \right)+3xyz+{{z}^{3}} $
 We know that $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}}+3x{{y}^{2}}+3{{x}^{2}}y={{\left( x+y \right)}^{3}} $ , so we can write the above equation as:
 $ \begin{align}
  & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}={{\left( x+y \right)}^{3}}-3xy\left( x+y+z \right)+3xyz+{{z}^{3}} \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}={{\left( x+y \right)}^{3}}+{{z}^{3}}-3xy\left( x+y+z \right)+3xyz \\
\end{align} $
Now, applying the formula $ {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{3}}+{{b}^{3}}-ab \right) $ , where $ a=x+y $ and $ b=z $ , we get:

 $ \begin{align}
  & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=\left( x+y+z \right)\{{{\left( x+y \right)}^{2}}+{{z}^{2}}-\left( x+y \right)z\}-3xy\left( x+y+z \right)+3xyz \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=\left( x+y+z \right)\{{{\left( x+y \right)}^{2}}+{{z}^{2}}-\left( x+y \right)z-3xy\}+3xyz \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+2xy+{{z}^{2}}-xz-yz-3xy \right)+3xyz \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=\left( x+y+z \right)\{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-\left( xy+yz+zx \right)\}+3xyz.........\left( 5 \right) \\
\end{align} $
On putting the value of xyz from equation (4) in equation (3), we get:
 $ xy+yz+zx=36\times \dfrac{2}{3}=24.........\left( 6 \right) $
On putting the values from equations (2), (4) and (6) in equation (5), we get:
 $ \begin{align}
  & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=12\left( 96-24 \right)+3\times \dfrac{2}{3} \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=12\times 72+2=864+2=866 \\
\end{align} $
Hence, the value of $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ is 866.

Note: A common mistake that students may do here is that they start finding the values of x, y and z from the given equations by solving them as there is three equations and three variables, so students may think that they can get the values of x, y and z by solving the three equations and then putting these values in the expression $ {{x}^{3}}+{{y}^{3}}+{{z}^{3}} $ to get the answer. But it would be a very lengthy approach and may not give the correct answer.