Question

don't give rubbish answer I will mark as a brain list for the correct answer​

Answer 1

Topic: Polynomials

Hint

• Polynomial identity

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$

• Equation

If the same number is multiplied, add, subtracted, or divided on both sides, the equation holds true(except for division by 0).

Solution

$\dfrac{x^2+1}{x}=3\dfrac{1}{3}$

$\implies \dfrac{x^2}{x} +\dfrac{1}{x} =\dfrac{10}{3}$

$\implies x+\dfrac{1}{x} =\dfrac{10}{3}$

Multiplying LHS by LHS, and RHS by RHS, (squaring on both sides)

$\implies (x+\dfrac{1}{x} )^2=\left(\dfrac{10}{3} \right)^2$

$\implies x^2+2+\dfrac{1}{x^2} =\dfrac{100}{9}$

$\implies x^2+\dfrac{1}{x^2} =\dfrac{82}{9}$

Subtracting 2 on both sides,

$\implies x^2-2+\dfrac{1}{x^2} =\dfrac{82}{9} -2$

$\implies \left(x-\dfrac{1}{x} \right)^2=\dfrac{64}{9}$

This quadratic equation gives $\boxed{x-\dfrac{1}{x} =\pm\frac{8}{3} }$.

Now multiplying LHS by LHS, RHS by RHS, three times, (cubing on both sides)

$\implies \left(x-\dfrac{1}{x} \right)^{3}=\pm\dfrac{512}{27}$

$\implies x^3-3\left(x-\dfrac{1}{x} \right)-\dfrac{1}{x^3} =\pm\dfrac{512}{27}$

$\implies x^{3}\mp3(\dfrac{8}{3} )-\dfrac{1}{x^3} =\pm\dfrac{512}{27}$

$\implies x^{3}-\dfrac{1}{x^3} =\pm\dfrac{512}{27}\pm8$

$\implies \boxed{x^3-\dfrac{1}{x^3} =\pm\dfrac{728}{27} }$

Answer 2

Variable, In algebra, a symbol (usually a letter) standing in for an unknown numerical value in an equation. Commonly used variables include x and y (real-number unknowns), z (complex-number unknowns), t (time), r (radius), and s (arc length).