If [3]a+[3]b+[3]c=0 then the value of (a+b+c/3)3 is equal to

Option a: abc. Solution: Let us define: x = [3]a, y = [3]b, z = [3]c According to the problem, we are given: x + y + z = 0 Recall the standard algebraic identity: if the sum of three variables is zero ( x + y + z = 0 ), then the sum of their cubes is equal to three times their product: x3 + y3 + z3 = 3xyz Substitute the definitions of x, y, z back into this identity: $ ([3]a)3 + ([3]b)3 + ([3]c)3 = 3 ([3]a × [3]b × [3]c) 重 a + b + c = 3 [3]abc Now, divide both sides of the equation by 3 : a + b + c/3 = [3]abc To find the final value, cube both sides of the equation: (a + b + c/3)3 = ([3]abc)3 = abc This matches Option A. Hence, Option A is the correct answer.

If $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0$ then the value of $\left(\frac{a+b+c}{3}\right)^3$ is equal to

The correct answer is Option a: $abc$. Let us define: $$x = \sqrt[3]{a}, \quad y = \sqrt[3]{b}, \quad z = \sqrt[3]{c}$$ According to the problem, we are given: $$x + y + z = 0$$ Recall the standard algebraic identity: if the sum of three variables is zero ($x + y + z = 0$), then the sum of their cubes is equal to three times their product: $$x^3 + y^3 + z^3 = 3xyz$$ Substitute the definitions of $x, y, z$ back into this identity: $$\left(\sqrt[3]{a}\right)^3 + \left(\sqrt[3]{b}\right)^3 + \left(\sqrt[3]{c}\right)^3 = 3 \left(\sqrt[3]{a} \cdot \sqrt[3]{b} \cdot \sqrt[3]{c}\right)$重 $$a + b + c = 3 \sqrt[3]{abc}$$ Now, divide both sides of the equation by $3$: $$\frac{a + b + c}{3} = \sqrt[3]{abc}$$ To find the final value, cube both sides of the equation: $$\left(\frac{a + b + c}{3}\right)^3 = \left(\sqrt[3]{abc}\right)^3 = abc$$ This matches **Option A**. Hence, **Option A** is the correct answer.

Ratio, Proportion, Indices, LogarithmMCQPYQ Jun 23Question 888 of 305

All Questions

If $\displaystyle \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0$ then the value of $\displaystyle \left(\frac{a+b+c}{3}\right)^3$ is equal to

Options

\displaystyle abc

\displaystyle 9abc

\displaystyle 1/(abc)

\displaystyle (1/9)abc

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — $\displaystyle abc$

All Options:

A $\displaystyle abc$
B $\displaystyle 9abc$
C $\displaystyle 1/(abc)$
D $\displaystyle (1/9)abc$

Detailed Solution & Explanation

Let us define:

x = \sqrt[3]{a}, \quad y = \sqrt[3]{b}, \quad z = \sqrt[3]{c}

According to the problem, we are given:

x + y + z = 0

Recall the standard algebraic identity: if the sum of three variables is zero (

\displaystyle x + y + z = 0

), then the sum of their cubes is equal to three times their product:

x^3 + y^3 + z^3 = 3xyz

Substitute the definitions of

\displaystyle x, y, z

back into this identity:

\displaystyle &#x27; in math mode at position 150: …rt[3]{c}\right)

̲重<br>" style="color:#cc0000">\left(\sqrt[3]{a}\right)^3 + \left(\sqrt[3]{b}\right)^3 + \left(\sqrt[3]{c}\right)^3 = 3 \left(\sqrt[3]{a} \cdot \sqrt[3]{b} \cdot \sqrt[3]{c}\right)

\displaystyle 重&lt;br&gt;</span>a + b + c = 3 \sqrt[3]{abc}<span class="katex-error" title="ParseError: KaTeX parse error: Can&#x27;t use function &#x27;

' in math mode at position 47: …he equation by

\displaystyle ̲3

:<br>" style="color:#cc0000"><br>Now, divide both sides of the equation by

\displaystyle 3

:<br>\frac{a + b + c}{3} = \sqrt[3]{abc}

<br>To find the final value, cube both sides of the equation:<br>

\left(\frac{a + b + c}{3}\right)^3 = \left(\sqrt[3]{abc}\right)^3 = abc$$
This matches **Option A**.
Hence, **Option A** is the correct answer.

About This Chapter: Ratio, Proportion, Indices, Logarithm

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

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If a3+b3+c3=0\displaystyle \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=03a​+3b​+3c​=0 then the value of (a+b+c3)3\displaystyle \left(\frac{a+b+c}{3}\right)^3(3a+b+c​)3 is equal to