Question

Prove that:
$(\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6}=970$

Answer 1

Answer:

Step-by-step explanation:

Hi,

Expanding (√3 + √2)⁶ using binomial expansion, we get

(√3 + √2)⁶ = ⁶C₀(√3)⁶ + ⁶C₁(√3)⁵(√2) + ⁶C₂(√3)⁴(√2)² + ⁶C₃(√3)³(√2)³

                + ⁶C₄(√3)²(√2)⁴ + ⁶C₅(√3)(√2)⁵ + ⁶C₆(√2)⁶---------------(1)

(√3 - √2)⁶ = ⁶C₀(√3)⁶ + ⁶C₁(√3)⁵(-√2) + ⁶C₂(√3)⁴(-√2)² + ⁶C₃(√3)³(-√2)³

                + ⁶C₄(√3)²(-√2)⁴ + ⁶C₅(√3)(-√2)⁵ + ⁶C₆(-√2)⁶

(√3 - √2)⁶ = ⁶C₀(√3)⁶ - ⁶C₁(√3)⁵(√2) + ⁶C₂(√3)⁴(√2)² - ⁶C₃(√3)³(√2)³

                + ⁶C₄(√3)²(√2)⁴ - ⁶C₅(√3)(√2)⁵ + ⁶C₆(√2)⁶---------------(2)

Adding (1) and (2), we get

(√3 + √2)⁶ + (√3 - √2)⁶

= 2*[⁶C₀(√3)⁶ + ⁶C₂(√3)⁴(√2)² + ⁶C₄(√3)²(√2)⁴ + ⁶C₆(√2)⁶]

=2*[27 + 15*9*2 + 15*3*4 + 8]

=2*[27 + 270 + 180 + 8]

=2*485

= 970

Hope, it helps !

Answer 2

Solution :

$(\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6}=970$

**********************************
$\rm\(x+a)^{n}=^nC_{0}x^{n}+^nC_{1}x^{n-1}a+^nC_{2}x^{n-2}a^{2}+....+^{n}C_{r}x^{n-r}a^{r}+...+^nC_{n}a^{n}$

**********************************
=> $\rm\displaystyle { [^6C_{0}^(\sqrt{3})^{6}+^{6}C_{1}(\sqrt {3})^{5}\(sqrt{2})^{1}+^{4}C_{2}(\sqrt{3})^{4}(\sqrt{2})^{2}+^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3}+{6}C_{4}(\sqrt{3})^{2}(\sqrt{2})^{4}+^{6}C_{5}(\sqrt{3})(\sqrt{2})^{5}+^{6}C_{6}(\sqrt{2})^{6}] - [^6C_{0}^(\sqrt{3})^{6}-^{6}C_{1}(\sqrt {3})^{5}(\sqrt{2})^{1}+^{4}C_{2}(\sqrt{3})^{4}(\sqrt{2})^{2}-^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3}+{6}C_{4}(\sqrt{3})^{2}(\sqrt{2})^{4}-^{6}C_{5}(\sqrt{3})(\sqrt{2})^{5}+^{6}C_{6}(\sqrt{2})^{6}] }$

= 2$[ ^{6}C_{0}(\sqrt{3})^{6}+^{6}C_{2}(\sqrt{3})^{4}(\sqrt{2})^{2}+^{6}C_{4}(\sqrt{3})^{2}(\sqrt{2})^{4}+^{6}C_{6}(\sqrt{2})^{6}$

= 2[1×3³+15×3²×2+15×3×4+1×2³]

= 2[ 27 + 270 + 180 + 18 ]

= 2 × 485

= 970

= RHS

••••