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Q13:
If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.

Solution :

Given Variables & Initial Setup

We are given the linear equation involving three variables:

$x + y + z = 0$

Objective: Prove the algebraic relationship $x^3 + y^3 + z^3 = 3xyz$.

To provide a comprehensive, masterclass-level proof, this relationship will be demonstrated using two distinct mathematical approaches: the Standard Identity Method and the Direct Algebraic Manipulation (Cubing) Method.


Method 1: Utilizing the Standard Algebraic Identity

Step 1: State the relevant polynomial identity

In algebra, the sum of three cubes is governed by the following fundamental identity:

$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$

[Theoretical Justification: This identity is derived by expanding the right-hand side and canceling out the intermediate cross-terms, leaving only the sum of the cubes and the $-3xyz$ term.]

Step 2: Substitute the given condition

We are given the premise that $x + y + z = 0$. We substitute this value directly into the right-hand side of our identity:

$x^3 + y^3 + z^3 - 3xyz = (0) \cdot (x^2 + y^2 + z^2 - xy - yz - zx)$

Step 3: Apply the Zero Product Property

According to the Zero Product Property, any finite real number or algebraic expression multiplied by zero results in zero. Therefore, the entire right-hand side collapses to $0$:

$x^3 + y^3 + z^3 - 3xyz = 0$

Step 4: Isolate the sum of the cubes

By adding $3xyz$ to both sides of the equation, we arrive at the final required expression:

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


Method 2: Direct Algebraic Manipulation (Cubing)

Step 1: Rearrange the initial equation

Starting with the given condition, isolate two variables on one side of the equation:

$x + y + z = 0 \implies x + y = -z$

Step 2: Cube both sides of the equation

To generate the cubic terms required for the proof, apply the power of 3 to both sides:

$(x + y)^3 = (-z)^3$

Step 3: Expand using the binomial cube identity

Expand the left side using the standard binomial identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$. Note that the cube of a negative value remains negative, so $(-z)^3 = -z^3$.

$x^3 + y^3 + 3xy(x + y) = -z^3$

Step 4: Substitute the initial rearranged condition

From Step 1, we established that $(x + y) = -z$. Substitute $-z$ back into the expanded equation in place of $(x + y)$:

$x^3 + y^3 + 3xy(-z) = -z^3$

Step 5: Simplify and rearrange terms

Multiply the terms to simplify the equation:

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

Finally, transpose $-z^3$ to the left side (becoming $+z^3$) and $-3xyz$ to the right side (becoming $+3xyz$):

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


Visual Representation of the Logical Pathways

The following flowchart illustrates the dual algebraic pathways utilized to prove the theorem, confirming the structural integrity of both methods.

x + y + z = 0 x + y = -z (x + y)³ = (-z)³ x³+y³+z³-3xyz = (0)(...) x³ + y³ + z³ = 3xyz Method 2 Method 1

Final Conclusion

Both the application of the standard cubic polynomial identity and direct algebraic expansion yield the exact same mathematical truth. When the sum of three variables is zero, the sum of their cubes is perfectly balanced by three times their product.

Final Solution: It is proven that if $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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