Question

Find (a+b)⁴ -(a-b)⁴ Hence evaluate (√3 + √2)⁴ - (√3-√2)⁴​

Answer 1

here is your answer

Step by Step explaination

Using binomial theorem the expressions (a + b)4 and (a – b)4 can be expressed as

(a + b)4 = 4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C3 ab3 +4C4 b4

(a – b)4 = 4C0 a4 – 4C1 a3b + 4C2 a2b2 – 4C3 ab3 +4C4 b4

(a + b)4 – (a – b)4 = [4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C3 ab3 +4C4 b4] – [4C0 a4 – 4C1 a3b + 4C2 a2b2 – 4C3 ab3 +4C4 b4]

= 2(4C1 a3b + 4C3ab3)

= 2(4a2b + 4ab3)

= 8ab(a2 + b2)

By putting a = √3 and b = √2, we obtain

(√3+√2)4 – (√3-√2)4 = 8(√3)( √2){(√3)2 + (√2)2}

= 8(√6){3 + 2}

= 40√6

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Answer 2

$\huge {\orange {\bf {\underline {\underline {Solutíon}}}:-}}$

we know,

$\sf (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³+ b⁴ ⠀⠀⠀⠀.....(1)$

$\sf (a - b)⁴ = a⁴ - 4a³b + 6a²b² - 4ab³+ b⁴ ⠀⠀⠀.....(2)$

Now, subtract (1) and (2)...

$\sf (a + b)⁴ - (a - b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ \\ \sf - a⁴ - 4a³b + 6a²b² - 4ab³ + b⁴ </p><p>(a + b)⁴ - (a - b)⁴$

$\sf = 2(4a³b + 4ab³)$

$\sf = 8ab(a² + b²) .......(3)$

substitute a = √3 and b = √2 in eq (3)

$\sf( \sqrt{3} + \sqrt{2} )^{2} - ( \sqrt{3} - \sqrt{2} )^{2} = 8 \sqrt{3} \sqrt{2} ( { \sqrt{2} }^{2} + \sqrt{3} ^{2} ) \\ \\ \sf= 8 \sqrt{6} (2 + 3) \\ \\ \sf = 8 \sqrt{6} \times 5 \\ \\ \sf = 40 \sqrt{6}$

Hence,

$\sf( \sqrt{3} + \sqrt{2} )^{2} - ( \sqrt{3} - \sqrt{2} )^{2} = 40 \sqrt{6 }$