Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem

Solution:

Using the binomial theorem, (3x² – 2ax + 3a²)³ can be expanded as follows by writing it as [(3x² + (– 2ax + 3a²)]³.

Then we get,

(3x² – 2ax + 3a²)³ = ³C₀ (3x²)³ + ³C₁ (3x²)² (-2ax + 3a²) + ³C₂ (3x²) (-2ax + 3a²)² + ³C₃ (-2ax + 3a²)³

Now using the formulas of (a+b)³ and (a+b)²,

= 27x⁶ + 3 (9x⁴) (-2ax + 3a²) + 3 (3x²) (4a²x² + 9a⁴ - 12 a³x) + (-8a³x³ + 36 a⁴x² - 54a⁵x + 27a⁶)

= 27x⁶ - 54ax⁵ + 81a²x⁴ + 36a²x⁴ + 81 a⁴x² - 108 a⁴x³ - 8 a³x³ + 36 a⁴x² - 54a⁵x + 27a⁶

=  27x⁶ - 54ax⁵ + 117a²x⁴ - 116a³x³ + 117a⁴x² - 54a⁵x + 27a⁶

NCERT Solutions Class 11 Maths Chapter 8 Exercise ME Question 10

Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem

Summary:

We found the expansion of (3x² – 2ax + 3a²)³ using binomial theorem to be 27x⁶ - 54ax⁵ + 117a²x⁴ - 116a³x³ + 117a⁴x² - 54a⁵x + 27a⁶