from a handpicked tutor in LIVE 1-to-1 classes

# 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,

(3x^{2} – 2ax + 3a^{2})^{3} = ^{3}C₀ (3x^{2})^{3} + ^{3}C₁ (3x^{2})^{2} (-2ax + 3a^{2}) + ^{3}C₂ (3x^{2}) (-2ax + 3a^{2})^{2} + ^{3}C₃ (-2ax + 3a^{2})^{3}

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

= 27x^{6} + 3 (9x^{4}) (-2ax + 3a^{2}) + 3 (3x^{2}) (4a^{2}x^{2} + 9a^{4} - 12 a^{3}x) + (-8a^{3}x^{3} + 36 a^{4}x^{2} - 54a^{5}x + 27a^{6})

= 27x^{6} - 54ax^{5} + 81a^{2}x^{4 }+ 36a^{2}x^{4} + 81 a^{4}x^{2} - 108 a^{4}x^{3} - 8 a^{3}x^{3} + 36 a^{4}x^{2} - 54a^{5}x + 27a^{6}

= 27x^{6} - 54ax^{5} + 117a^{2}x^{4} - 116a^{3}x^{3} + 117a^{4}x^{2 }- 54a^{5}x + 27a^{6}

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 27x^{6} - 54ax^{5} + 117a^{2}x^{4} - 116a^{3}x^{3} + 117a^{4}x^{2 }- 54a^{5}x + 27a^{6}

visual curriculum