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If ${x^a}.{x^b}.{x^c} = 1{\text{ then }}{a^3} + {b^3} + {c^3}$ is equal to
(A). 9
(B). $abc$
(C). $a + b + c$
(D).$3abc$

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Last updated date: 17th Apr 2024
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Answer
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Hint- In order to solve this question, take log in both sides of the first expression to find the value of $a + b + c$ and then by using the formula given as ${\left( {a + b + c} \right)^3} = \left( {{a^3} + {b^3} + {c^3}} \right) + 3\left( {\left( {a + b + c} \right)\left( {ab + bc + ca} \right) - abc} \right)$ we will proceed further.

Complete step by step answer:
Given equation ${x^a}.{x^b}.{x^c} = 1.$
We have to find ${a^3} + {b^3} + {c^3}$
As we know that ${z^p}.{z^q}.{z^r} = {z^{p + q + r}}{\text{ }}$
So by using it in given equation, we get
${x^{a + b + c}} = 1$
Now, by taking log to both sides, we get
$
   \Rightarrow \log {x^{a + b + c}} = \log 1 \\
   \Rightarrow \left( {a + b + c} \right)\log x = 0{\text{ }}\left[ {\because \log {x^p} = p\log x{\text{ and }}\log 1 = 0} \right] \\
  {\text{either }}\left( {a + b + c} \right) = 0{\text{ or }}\log x = 0 \\
$
Now, we will use the formula of ${\left( {a + b + c} \right)^3}$ which is given as
${\left( {a + b + c} \right)^3} = \left( {{a^3} + {b^3} + {c^3}} \right) + 3\left( {\left( {a + b + c} \right)\left( {ab + bc + ca} \right) - abc} \right)$
Substituting the value of $\left( {a + b + c} \right) = 0$ we get
\[ \Rightarrow 0 = \left( {{a^3} + {b^3} + {c^3}} \right) + 3\left( {0 \times \left( {ab + bc + ca} \right) - abc} \right)\]
By simplifying the above equation, we ge
\[
   \Rightarrow 0 = \left( {{a^3} + {b^3} + {c^3}} \right) + 3\left( { - abc} \right) \\
   \Rightarrow {a^3} + {b^3} + {c^3} = 3abc \\
\]
Hence, the value of \[{a^3} + {b^3} + {c^3} = 3abc\] and the correct answer is “D”.

Note- In order to solve these types of questions, first of all remember all the algebraic identities and you must be aware of how to solve linear algebraic equations and have knowledge of terms like variables. In the above question we have also used logarithmic function properties. So, you must have a good knowledge of logarithm and exponents.