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
244.8k+ 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
NEET UG Exam Countdown 2026 – Days Left, Tracker & Tips

JEE Main 2026 Admit Card OUT LIVE Soon| Session 2 Direct Download Link

JEE Main 2026 Session 2 City Intimation Slip Expected Soon: Check How to Download

JEE Main 2026 Session 2 Application Form: Reopened Registration, Dates & Fees

JEE Main 2026 Session 2 Registration (Reopened): Last Date, Fees, Link & Process

WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

Understanding the Block and Tackle System

Derivation of Equation of Trajectory Explained for Students

Understanding Average and RMS Value in Electrical Circuits

