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If (a3 + b+ c3 - 3abc) = 405, and (a - b)2 + (b - c)2 + (c -a)2 = 54, find the value of (a + b + c).

This question was previously asked in
SSC CGL 2023 (Tier-II) Paper 1 Previous Year Paper (26-Oct-2023) (Shift-1)
The correct answer is

15

Solving for \(a+b+c\) Using Algebraic Identities

This problem requires us to find the value of \((a + b + c)\) using the given values for two algebraic expressions involving \(a\), \(b\), and \(c\). We are given:

  • \(a^3 + b^3 + c^3 - 3abc = 405\)
  • \((a - b)^2 + (b - c)^2 + (c - a)^2 = 54\)

Key Algebraic Identities

To solve this problem, we need to recall two important algebraic identities:

  1. The identity for the sum of cubes minus three times the product:

    \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)

  2. The identity relating the sum of squares of differences:

    \((a - b)^2 + (b - c)^2 + (c - a)^2\)

    Let's expand this expression:

    \((a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ca + a^2)\)

    \(= a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2\)

    \(= 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca\)

    \(= 2(a^2 + b^2 + c^2 - ab - bc - ca)\)

    So, \((a - b)^2 + (b - c)^2 + (c - a)^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca)\).

Connecting the Given Expressions

Notice that the term \((a^2 + b^2 + c^2 - ab - bc - ca)\) appears in both identities. From the second identity, we can write:

\(a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2} \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right)\)

Now, substitute this into the first identity:

\(a^3 + b^3 + c^3 - 3abc = (a + b + c) \times \frac{1}{2} \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right)\)

Substituting the Given Values and Solving

We are given the values for both sides of the equation on the right:

  • \(a^3 + b^3 + c^3 - 3abc = 405\)
  • \((a - b)^2 + (b - c)^2 + (c - a)^2 = 54\)

Substitute these values into the combined identity:

\(405 = (a + b + c) \times \frac{1}{2} \times 54\)

\(405 = (a + b + c) \times 27\)

To find the value of \((a + b + c)\), divide both sides by 27:

\(a + b + c = \frac{405}{27}\)

Performing the division:

\(405 \div 27 = 15\)

So, the value of \((a + b + c)\) is 15.

The final answer is 15.

Revision Table: Algebraic Identities Used

Identity Name Formula
Sum of Cubes Identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
Sum of Squares of Differences \((a - b)^2 + (b - c)^2 + (c - a)^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca)\)

Additional Information on Related Algebraic Concepts

Algebraic identities are equations that are true for all values of the variables involved. They are powerful tools for simplifying expressions, solving equations, and proving other mathematical relationships.

The identity \(a^3 + b^3 + c^3 - 3abc\) is particularly useful and has interesting properties. For instance, if \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).

The expression \((a - b)^2 + (b - c)^2 + (c - a)^2\) represents twice the sum of the squared differences between variables taken pairwise. It is always non-negative, and it is zero if and only if \(a = b = c\).

Understanding and memorizing common algebraic identities significantly helps in solving complex algebraic problems efficiently during exams.

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Important Questions from Algebra

  1. The difference between the two positive numbers x and y where x > y, is 25% of x. If the value of y is 15, then the value of x is:

  2. If p2 + q2 - r2 = 0, then the value of (p6 + q6 - r6) ÷ p2q2r2 is:

  3. If √2 + √x = √3, then the value of x is equal to:

  4. The sum of two numbers is 20 and their difference is 2.5. Ratio of these numbers will be:

  5. If \(\rm \frac{\sqrt{19 - x \sqrt{12}}}{1} = \sqrt 4 - \sqrt 3\)  then the value of x is equal to:

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