The expression x^4 + 1/x^4 + 1 can be factorized as (x^2 + 1/x2 - 2. The expression (x^2 + 1/x2 - 2 can be further factorized as (x^2 - 1/x^2 + 1)(x^2 + 1/x^2 + 1)
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kridaytheking
To factorize the expression x^4 + 1/x^4 + 1, we can treat it as a quadratic expression in x^4. Let's substitute y = x^4:
y^2 + 1/y + 1
Now, we have a quadratic expression in y. To factorize this quadratic expression, we can use the formula for factoring a quadratic equation of the form ay^2 + by + c:
y^2 + 1/y + 1 = (y + y_1)(y + y_2)
where y_1 and y_2 are the roots of the quadratic equation.
To find the roots, we can use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 1, and c = 1:
y = (-(1) ± √(1 - 4(1)(1))) / 2(1) y = (-1 ± √(-3)) / 2
Since the discriminant (√(b^2 - 4ac)) is negative, the roots are complex. Let's write them in the form a + bi:
y_1 = -1/2 + (i√3)/2 y_2 = -1/2 - (i√3)/2
Now, let's rewrite the expression factored in terms of y:
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The expression x^4 + 1/x^4 + 1 can be factorized as (x^2 + 1/x2 - 2. The expression (x^2 + 1/x2 - 2 can be further factorized as (x^2 - 1/x^2 + 1)(x^2 + 1/x^2 + 1)
your body is nice
y^2 + 1/y + 1
Now, we have a quadratic expression in y. To factorize this quadratic expression, we can use the formula for factoring a quadratic equation of the form ay^2 + by + c:
y^2 + 1/y + 1 = (y + y_1)(y + y_2)
where y_1 and y_2 are the roots of the quadratic equation.
To find the roots, we can use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 1, and c = 1:
y = (-(1) ± √(1 - 4(1)(1))) / 2(1)
y = (-1 ± √(-3)) / 2
Since the discriminant (√(b^2 - 4ac)) is negative, the roots are complex. Let's write them in the form a + bi:
y_1 = -1/2 + (i√3)/2
y_2 = -1/2 - (i√3)/2
Now, let's rewrite the expression factored in terms of y:
y^2 + 1/y + 1 = (y + y_1)(y + y_2)
Replace y with x^4:
(x^4 - 1/2 + (i√3)/2)(x^4 - 1/2 - (i√3)/2)
So, the factored form of x^4 + 1/x^4 + 1 is:
(x^4 - 1/2 + (i√3)/2)(x^4 - 1/2 - (i√3)/2)