default_background

Find the best tutors and institutes for Class 10 Tuition

Find Best Class 10 Tuition

Please select a Category.

Please select a Locality.

No matching category found.

No matching Locality found.

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


Top Tutors who teach Polynomials

(203)
Top Tutor
locationImg Venkatraman Nagar Chennai, Chennai
experience 10 yrs of Exp students 27 students
greenClock 3790 hrs of classes completed
rsIcon 320 per hour
Expereinced Tutor for class 9- class 12 on maths and sxience subjects
Online Classes Online Classes
regularClasses Tutor's home

im having good experince in teaching with make them understand the concepts better

Amudha

Classes are much interactive so useful to assess the students understanding capacity during the class itself.

Skills: Mathematics , Science
Also teaches: Class 11 Tuition, Class 9 Tuition and more

(32)
Top Tutor
locationImg Electronic City Electronics City Phase 1, Bangalore
experience 15 yrs of Exp students 5 students
greenClock 39 hrs of classes completed
rsIcon 1000 per hour
Teacher With 15 Years of Experience
Online Classes Online Classes

Ph.D enrolled, M.Tech(Electrical eng). Over 15 years of experience in teaching at different Engineering colleges and tutoring across different boards including CBSE, ICSE, IGCSE, IBDP, Level AS/A in Mathematics & Science. Approachable to students & succeeded to boost their fundamental knowledge.

Akhil

Ms. Dipika always ensures that students are thorough with the concepts and discourages mugging the formulae. She is chill with students; she actively helps and encourages students to perform their best rather than nagging them about a poor performance. Her teaching methods are simple, effective and logical. It helps build a strong conceptual foundation for advanced topics. Overall, Ms. Dipika is the perfect Maths tutor!

Skills: Mathematics
Also teaches: Class 6 Tuition, Class 9 Tuition and more

Deepak Joshi Class 10 Tuition trainer in Bangalore
(48)
Top Tutor
locationImg Ashok Nagar D' Souza Layout, Bangalore
experience 10 yrs of Exp students 25 students
greenClock 770 hrs of classes completed
rsIcon 600 per hour
Ph.D. Candidate | M.Sc. (Mathematics) | Former Faculty at BYJU'S (7 Years) & DPS Gurgaon (3 Years)
Online Classes Online Classes

10+ years of experience teaching Mathematics for Grades 10 to 12th • Consistent record of 100% academic results • Very friendly and supportive learning environment • Encourages students to ask questions freely and confidently • Strong focus on conceptual clarity and problem-solving skills • Helps students build confidence and excel in Mathematics • Fee Structure for One-to-One Classes: ₹700 per hour • Group Classes: ₹4,000 per month.

Pratima Dixit

Sir clears all my doubts he makes difficult concepts easy to understand he takes Previous year questions which makes it easier to prepare

Skills: Mathematics
Also teaches: Class 9 Tuition, Class 11 Tuition and more

locationImg Indirapuram, Ghaziabad
experience 8 yrs of Exp students 3 students
rsIcon 300 per hour
An tutor with 10year+ experience
Online Classes Online Classes
homeTutions Student's Home

I am a teacher of maths and science, having a experience of 8+ years in teaching, done my schooling from CBSe board and degree from IP university.

Skills: Mathematics , Science
Also teaches: Class 11 Tuition, Class 8 Tuition and more

(11)
Top Tutor
locationImg Beta I Block E, Noida
experience 5 yrs of Exp students 1 student
rsIcon 400 per hour
Best Education at best place unifyclasses.
Online Classes Online Classes
homeTutions Student's Home
regularClasses Tutor's home

Dear Parents and students, I am Amit, We provide home-based tuition for subjects and we also do professional coaching for NEET/IIT-JEE and other competitive exams. We have experienced tutors in all the fields and subjects. We also provide demo class free of cost. We do online as well as offline mode of teaching...

Skills: Mathematics , Physics and more

Find more Tutor for Polynomials in your City


Other Subjects in CBSE - Class 9

Worksheet Icon

Download free CBSE - Class 9 Mathematics Polynomials EXERCISE 2.4 worksheets

Download Now

Find Best Class 10 Tuition ?

Find Now »