Question

If x^2+1/x^2=7find 7x^3+8^x-7/x^3-8/x

Answer 1

Given :-

• x² + 1/x² = 7

To Find :-

• 7x³ + 8x - (7/x³) - (8/x) = ?

Solution :-

→ 7x³ + 8x - (7/x³) - (8/x)

Re - arranging ,

→ (7x³ - 7/x³) + (8x - 8/x)

→ 7(x³ - 1/x³) + 8(x - 1/x)

using a³ - b³ = (a - b)(a² + b² + ab) Now,

→ 7[(x - 1/x)(x² + 1/x² + 1)] + 8(x - 1/x)

Taking (x - 1/x) common Now,

→ (x - 1/x) [ 7(x² + 1/x² + 1) + 8 ]

Putting given value of x² + 1/x² now, we get,

→ (x - 1/x) [ 7( 7 + 1) + 8 ]

→ (x - 1/x) [ 7 * 8 + 8 ]

→ (x - 1/x) [ 56 + 8 ]

→ (x - 1/x) * 64 --------------- Equation (1).

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Now,

→ x² + 1/x² = 7

Subtracting 2 Both Sides we get,

→ x² + 1/x² - 2 = 7 - 2

Comparing LHS with a² + b² - 2ab = (a - b)² , we get,

→ (x - 1/x)² = 5

Square - Root Both sides Now,

→ (x - 1/x) = ± √5 --------------- Equation (2).

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Putting Equation (2) in Equation (1) Now , we get,

→ Ans :- 64√5 or (-64√5) .

Answer 2

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$\huge\tt{GIVEN:}$

• x²+1/x²=7

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$\huge\tt{TO~FIND:}$

• 7x^3+8^x-7/x^3-8/x

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$\huge\tt{SOLVING:}$

➩7x³+8x - (7/x³) -(8/x)

➩(7x³-7/x³)+(8x-8/x)

➩7(x³-1/x³)+8(x-1/x)

➩7[(x-1/x)(x²+1/x²+1)] + 8(x-1/x)

➩(x-1/x)[7(x²+1/x²+1)+8]

➩(x-1/x)[7(7+1)+8]

➩(x-1/8)[56+8]

➩(x-1/x)×64 ____(EQ.1)

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➩x²+1/x² =7

➩x²+1/x²-2 = 7-2

➩(x-1/x²)=5

➩(x-1/x)=±√5 ______(EQ.2)

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By comparing both equations,

➠64√5 or (-64√5)

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