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$\tt\large\underline\green{Question:-}$
$Factorise:\ \textless \ br /\ \textgreater \ [tex] {x}^{2} + \frac{1}{ {x}^{2} } - 7(x - \frac{1}{x} ) + 8\ \textless \ br /\ \textgreater \ Hint - Put (x-1/x)= y\ \textless \ br /\ \textgreater \ ​$

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Answer 1

Step-by-step explanation:

Given :-

[X²+(1/X²)] -7[X-(1/X)] +8

To find :-

Factorization of the expression

Solution :-

Given expression is

[X²+(1/X²)] -7[X-(1/X)] +8

We know that

(a-b)² = a²+b²-2ab

=> [{X-(1/X)}²+2(X)(1/X)] -7[X-(1/X)]+8

Where, a = X and b = 1/X

=> [{X-(1/X)}²+2(X/X)] -7[X-(1/X)]+8

=> [{X-(1/X)}²+2(1)] -7[X-(1/X)]+8

=> [{X-(1/X)}²+2] -7[X-(1/X)]+8

=> [X-(1/X)]²+2 -7[X-(1/X)]+8

=> [X-(1/X)]² -7[X-(1/X)]+(8+2)

=> [X-(1/X)]² -7[X-(1/X)]+10

=> [X-(1/X)]² -5[X-(1/X)]-2[X-(1/X)]+10

=> [X-(1/X)][{X-(1/X))-5]-2[{X-(1/X)}-5]

=> [{X-(1/X)}-2][{X-(1/X)-5]

Alternative Method:-

Given expression is

[X²+(1/X²)] -7[X-(1/X)] +8

We know that

(a-b)² = a²+b²-2ab

=> [{X-(1/X)}²+2(X)(1/X)] -7[X-(1/X)]+8

Where, a = X and b = 1/X

=> [{X-(1/X)}²+2(X/X)] -7[X-(1/X)]+8

=> [{X-(1/X)}²+2(1)] -7[X-(1/X)]+8

=> [{X-(1/X)}²+2] -7[X-(1/X)]+8

Put X-(1/X) = Y then

=> Y²+2-7Y+8

=> Y²-7Y+10

=> Y²-5Y-2Y+10

=> Y(Y-5)-2(Y-5)

=> (Y-2)(Y-5)

=> [{X-(1/X)}-2][{X-(1/X)-5]

Answer :-

[X²+(1/X²)]-7[X-(1/X)]+8

=[{X-(1/X)}-2][{X-(1/X)-5]

Used formulae:-

→ (a-b)² = a²+b²-2ab

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given :-

[X²+(1/X²)] -7[X-(1/X)] +8

To find :-

Factorization of the expression

Solution :-

Given expression is

[X²+(1/X²)] -7[X-(1/X)] +8

We know that

(a-b)² = a²+b²-2ab

=> [{X-(1/X)}²+2(X)(1/X)] -7[X-(1/X)]+8

Where, a = X and b = 1/X

=> [{X-(1/X)}²+2(X/X)] -7[X-(1/X)]+8

=> [{X-(1/X)}²+2(1)] -7[X-(1/X)]+8

=> [{X-(1/X)}²+2] -7[X-(1/X)]+8

=> [X-(1/X)]²+2 -7[X-(1/X)]+8

=> [X-(1/X)]² -7[X-(1/X)]+(8+2)

=> [X-(1/X)]² -7[X-(1/X)]+10

=> [X-(1/X)]² -5[X-(1/X)]-2[X-(1/X)]+10

=> [X-(1/X)][{X-(1/X))-5]-2[{X-(1/X)}-5]

=> [{X-(1/X)}-2][{X-(1/X)-5]

$answer}$