
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).$3abc$
Answer
232.8k+ views
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.
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.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

