Let x = √(4-√7) and y = √(4+√7), then value of (x-y/√3)² + (x²y+xy²)²+(x+y)² is -​

Given info:- X = √(4-√7) and Y = √(4+√7)

Find the value of [(X-Y)/√3]²+(X²Y+XY²)²+(X+Y)²?

Explanation:-

Given that:

X = √(4-√7)

It can be written as

On multiplying and dividing by 2 in the squre root

⇛ X = √[2{(4-√7)}/2]

⇛ X =√[(8-2√7)/2]

⇛ X =√[{(7+1)-2√7{/2]

⇛ X =√[{(√7)²+(√1)²-2(√7)(√1)}/2]

It is in the form of a²-2ab+b²

Where , a = √7 and b = √1

⇛X = √[{(√7-√1)}²/2] [∵(a-b)² = a²-2ab+b²]

⇛ X = (√7-√1)/√2

⇛ X = (√7-1)/√2 — — — — Eqⁿ(1)

and

Y = √(4+√7)

It can be written as

On multiplying and dividing by 2 in the squre root

⇛ Y = √[{2(4+√7)}/2]

⇛ Y =√[(8+2√7)/2]

⇛ Y =√[{(7+1)+2√7}/2]

⇛ Y =√[{(√7)²+(√1)²+2(√7)(√1)}/2]

It is in the form of a²+2ab+b²

Where , a = √7 and b = √1

⇛Y = √[{(√7+√1)}²/2] [∵(a+b)² = a²+2ab+b²]

⇛ Y = (√7+√1)/√2

⇛ Y = (√7+1 )/√2 — — — — Eqⁿ(2)

On adding Eqⁿ (1) & Eqⁿ(2)

→ X+Y = (√7-1)/√2+(√7+1)/√2

→ X+Y = (√7-1+√7+1)/√2

→ X+Y = (2√7 )/√2

→ X+Y = (√2×√2×√7)/√2

→ X+Y = √2×√7

→ X+Y = √14 — — — — Eqⁿ(3)

On Subtracting Eqⁿ(2) from Eqⁿ(1)

→ X-Y =[(√7-1)/√2]-[(√7+1)/√2]

→ X-Y =[ (√7-1 )-(√7+1)]/√2

→ X-Y = (√7-1-√7-1)/√2

→ X-Y = (-1-1)/√2

→ X-Y = -2/√2

→ X-Y = -(√2×√2)/√2

→ X-Y = -√2 — — — — Eqⁿ(4)

On multiplying Eqⁿ(1)& Eqⁿ(2)

→ XY = [(√7-1)/√2][(√7+1)/√2]

→ XY = (√7-1)×(√7+1)/(√2×√2)

→ XY = [(√7)²-(1)²]/(2) [∵(a+b)(a-b) = a²-b²]

→ XY = (7-1)/2

→ XY = 6/2

→ XY = 3 — — — — — Eqⁿ(5)

Now,

The value of [(X-Y)/√3]²+(X²Y+XY²)²+(X+Y)²

It can be written as

⇛[(X-Y)²/3] + [ XY(X+Y) ]²+(X+Y)²

From Eqⁿ(3), Eqⁿ(4)& Eqⁿ(5)

⇛ [(-√2)²/3] + [ 3(√14)]²+(√14)²

⇛ (2/3) + (9×14)+(14)

⇛ (2/3)+(126)+(14)

⇛ (2/3)+140

⇛ [2+(3×140)]/3

⇛(2+420)/3

⇛ 422/3

Therefore, the value of [(X-Y)/√3]²+(X²Y+XY²)²+(X+Y)² is 422/3.

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