Question

Evaluate $ {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} $ without calculating the cubes.

Answer 1

Hint: In the given problem, to find required value first we will rewrite the given expression. Note that $ - {\left( a \right)^3} $ can be written as $ {\left( { - a} \right)^3} $ because the power is odd. If $ a + b + c = 0 $ then $ {a^3} + {b^3} + {c^3} $ is equal to $ 3abc $ . We will use this result to find the required value.

Complete step-by-step answer:
In this problem, we have to evaluate $ {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} \cdots \cdots \left( 1 \right) $ . We know that $ - {\left( a \right)^3} $ can be written as $ {\left( { - a} \right)^3} $ . Using this information let us rewrite the expression $ \left( 1 \right) $ . So, we can write $ {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} + {\left( { - \dfrac{5}{6}} \right)^3} \cdots \cdots \left( 2 \right) $ .
Let us take $ a = \dfrac{1}{2},b = \dfrac{1}{3} $ and $ c = - \dfrac{5}{6} $ . Now we are going to find $ a + b + c $ . So, we can write
 $ a + b + c = \dfrac{1}{2} + \dfrac{1}{3} + \left( { - \dfrac{5}{6}} \right) $
 $ \Rightarrow a + b + c = \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{5}{6} $
Let us simplify the RHS of the above equation by taking LCM. Note that LCM of numbers $ 2,3,6 $ is $ 6 $ . So, we can write
 $ \Rightarrow a + b + c = \dfrac{3}{6} + \dfrac{2}{6} - \dfrac{5}{6} $
 $ \Rightarrow a + b + c = \dfrac{{3 + 2 - 5}}{6} $
 $ \Rightarrow a + b + c = 0 $
Now we have $ a + b + c = 0 $ where $ a = \dfrac{1}{2},b = \dfrac{1}{3} $ and $ c = - \dfrac{5}{6} $ . We have to find the value of $ {a^3} + {b^3} + {c^3} $ where $ a = \dfrac{1}{2},b = \dfrac{1}{3} $ and $ c = - \dfrac{5}{6} $ . If $ a + b + c = 0 $ then $ {a^3} + {b^3} + {c^3} $ is equal to $ 3abc $ . Now we are going to use this result. So, we can write
 $ {a^3} + {b^3} + {c^3} = 3abc $
 $ \Rightarrow {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} = 3\left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{3}} \right)\left( { - \dfrac{5}{6}} \right) $
 $ \Rightarrow {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} = - \dfrac{5}{{12}} $
Hence, the required value is $ - \dfrac{5}{{12}} $ .

Note: In the given problem, it is mentioned that we need to evaluate the value without calculating the cubes. If it is not mentioned then we can evaluate the value by taking cubes of individual terms. That is, we can write
 $ {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} = \dfrac{1}{8} + \dfrac{1}{{27}} - \dfrac{{125}}{{216}} $
Simplify the above expression by taking LCM. Note that LCM of numbers $ 8,27,216 $ is $ 216 $ . So, we can write
 $ {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} = \dfrac{{27}}{{216}} + \dfrac{8}{{216}} - \dfrac{{125}}{{216}} $
 $ \Rightarrow {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} = - \dfrac{{90}}{{216}} = - \dfrac{5}{{12}} $
Also remember that if $ a = b = c $ then we can write $ {a^3} + {b^3} + {c^3} $ is equal to $ 3abc $ . Factorization of $ {a^3} + {b^3} + {c^3} - 3abc $ is given by $ \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $ .