Question

Solve with proper explanation:
(sqrt 2 + sqrt 3)^3 - (1/(sqrt 2 + sqrt3)^3)

Answer 1

Solution :-

Given that

[(√2+√3)³ ] - [1/(√2+√3)³]

It can be written as

= [ (√2+√3)³ ] - [ 1/(√2+√3)]³

On taking 1/(√2+√3)

The denominator = √2+√3

The Rationalising factor of √a+√b is √a-√b

The Rationalising factor of √2+√3 is √2-√3

On Rationalising the denominator then

= [/1(√2+√3)]×[(√2-√3)/(√2-√3)]

= (√2-√3)/[(√2+√3)(√2-√3)]

= (√2-√3)/([(√2)²-(√3)²]

Since, (a+b)(a-b) = a²-b²

Where , a = √3 and b = √2

= (√2-√3)/(2-3)

= (√2-√3)/(-1)

= -(√2-√3)

= -√2+√3

= √3-√2

Therefore, 1/(√2+√3) = √3-√2

Now,

Given expression becomes

(√2+√3)³-(√3-√2)³

= [(√2)³+(√3)³+3(√2)²(√3)+3(√2)(√3)²] - [(√3)³-(√2)³ -3(√3)²(√2) +3(√3)(√2)²]

Since, (a+b)³ = a³+b³+3a²b+3ab²

and (a-b)³ = a³-b³-3a²b+3ab²

= [(√2)³+(√3)³+3(2)(√3)+3(√2)(3)] - [(√3)³-(√2)³ -3(3)√2)+3(√3)(2)]

= [(√2)³+(√3)³+6√3+9√2] - [(√3)³-(√2)²-9√2+6√3]

= (√2)³+(√3)³+6√3+9√2-(√3)³+(√2)³+ 9√2 - 6√3

= [(√2)³+(√2)³]+[(√3)³-(√3)³]+[6√3-6√3] + (9√2+9√2)

= (√)2³+(√2)³+0+0+9√2+9√2

= (√2)³+(√2)³+9√2+9√2

= 2(√2)³+(9+9)√2

= 2(√2)³+18√2

= 2(√2×√2×√2)+18√2

= 2(2√2)+18√2

= 4√2+18√2

= (4+18)√2

= 22√2

Answer :-

[(√2+√3)³ ] - [1/(√2+√3)³] = 22√2

Used formulae:-

• The Rationalising factor of √a+√b is √a-√b

• (a+b)³ = a³+b³+3a²b+3ab²

• (a-b)³ = a³-b³-3a²b+3ab²

• (a+b)(a-b) = a²-b²

Answer 2

$\bold{ANSWER≈}$

Given that

[(√2+√3)³ ]-[1/(√2+√3)³]

It can be written as

= [(√2+√3)³ ]-[1/(√2+√3)]³

On taking 1/(√2+√3)

The denominator = √2+√3

The Rationalising factor of Va+√b is va-vb

The Rationalising factor of √2+√3 is

√2-√3

On Rationalising the denominator then

= [/1(√2+√3)]*[(√2-√3)/(√2-√3)]

=(√2-√3)/[(√2+√3)(√2-√3)]

=(√2-√3)/([(√2)²-(√3)²]

Since, (a+b)(a-b) = a²-b²

Where, a = √3 and b = √2

=(√2-√3)/(2-3)

=(√2-√3)/(-1)

= -(√2-√3)

= -√2+√3

= √3-√2

Therefore, 1/(√2+√3)=√3-√2

Now,

Given expression becomes

(√2+√3)³-(√3-√2)³

= [(√2)³+(√3)³+3(√2)²(√3)+3(√2)(√3)²] - [(√3)³-(√2)³ -3(√3)²(√2) +3(√3)(√2)²]

Since, (a+b)³ = a³+b³+3a²b+3ab²

and (a-b)³ a³-b³-3a²b+3ab²

= [(√2)³+(√3)³+3(2)(√3)+3(√2)(3)] -

[(√3)³-(√2)³-3(3)√2)+3(√3)(2)]

= [(√2)³+(√3)³+6√3+9√2] -

[(√3)³-(√2)²-9-√2+6√3]

=(√2)³+(√3)³ +6√3+9√2-(√3)³+(√2)³+ 9-√2

- 6-√3

= [(√2)³+(√2)³]+[(√3)³-(√3)³]+[6√3-6√3] +

(9-√2+9√2)

=(√)2³+(√2)³+0+0+9√2+9√2

=(√2)³+(√2)³3+9√2+9-√2

= 2(√2)³+(9+9)√2

= 2(√2)³+18-√2

2(√2x√2x√2)+18-√2

= 2(2-√2)+18-√2

=

= 4√2+18-√2

= (4+18) -√2

= 22√2

Answer :

[(√2+√3)³] - [1/(√2+√3)³] = 22√2

Used formulae:

• The Rationalising factor of Va+√b is

Va-vb

• (a+b)³ = a³+b³+3a²b+3ab²

(a-b)³ = a³-b³-3a²b+3ab²

• (a+b)(a-b) = a²-b²