If a, b, c are all non-zero and a + b + c = 0, prove that a²/bc + b²/ca + c²/ab = 3.

Solution:

Given, a, b and c are all non-zero

Also, a + b + c = 0

We have to prove that a²/bc + b²/ca + c²/ab = 3.

Using the algebraic identity,

x³ + y³ + z³ - 3xyz = (x + y + z) (x² +y² + z² - xy - yz - zx)

If x + y + z = 0, then x³ + y³ + z³ - 3xyz = 0

So, x³ + y³ + z³ = 3xyz.

Given, a + b + c = 0

So, a³ + b³ + c³ = 3abc

Dividing both sides by 3abc,

a³/3abc + b³/3abc + c³/3abc = 3abc/3abc

a²/3bc + b²/3ca + c²/3ab = 1

a²/bc + b²/ca + c²/ab = 3

Hence proved.

✦ Try This: Find the value of k if (x-2)is a factor of polynomial p(x) = 2x³ - 6x² + 5x + k.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2

NCERT Exemplar Class 9 Maths Exercise 2.4 Problem 7

If a, b, c are all non-zero and a + b + c = 0, prove that a²/bc + b²/ca + c²/ab = 3

Summary:

If a, b, c are all non-zero and a + b + c = 0, using the algebraic identity it is proven that a²/bc + b²/ca + c²/ab = 3

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