X⁴+1/x⁴=47,find the value of x³+1/x³

The value of x^3 + 1/x^3 is 47/x + 1/x^3 - 1/x^5

Step-by-step explanation:

x^4 + 1/x^4 = 47

x^4 = 47 - 1/x^4

x^3 + 1/x^3 = 1/x(x^4 + 1/x^2)

x^4 = 47 - 1/x^4

x^3 + 1/x^3 = 1/x(47 - 1/x^4 + 1/x^2) = 47/x - 1/x^5 + 1/x^3 = 47/x + 1/x^3 - 1/x^5

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

answer: 43/157

this is the answer and if you try to prove me wrong your a meanie