# Prove that<

Let Δ =

Applying Elementary Transformations

C1 C1 + C2 + C3, we have

Δ =

Taking (a + b + c) common from C1, we get

Δ = (a + b + c)

Applying R2 R2 – R1 and R3 R3 – R1

Expanding along C1

Δ = (a + b + c) [1 × {(2b + a) (2c + a) – (a – b) (a – c)} – 0 + 0]

Δ = (a + b + c) [4bc + 2ab + 2ac + a2 – a2 + ac + ba – bc]

Δ = (a + b + c) (3ab + 3bc + 3ac)

Δ = 3 (a + b + c) (ab + bc + ca)

Hence, the given result proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Prove the followiMathematics - Board Papers