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Exercise

$\left(z^2+\frac{1}{z^2}\right)^4$

Step-by-step Solution

1

Expand the binomial $\left(z^2+\frac{1}{z^2}\right)^4$

$z^{8}+\frac{4z^{6}}{z^2}+6z^{4}\left(\frac{1}{z^2}\right)^2+4z^2\left(\frac{1}{z^2}\right)^3+\left(\frac{1}{z^2}\right)^4$
2

Simplify the fraction $\frac{4z^{6}}{z^2}$ by $z$

$z^{8}+4z^{4}+6z^{4}\left(\frac{1}{z^2}\right)^2+4z^2\left(\frac{1}{z^2}\right)^3+\left(\frac{1}{z^2}\right)^4$
3

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$z^{8}+4z^{4}+6z^{4}\left(\frac{1}{z^{4}}\right)+4z^2\left(\frac{1}{z^2}\right)^3+\left(\frac{1}{z^2}\right)^4$
4

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$z^{8}+4z^{4}+6z^{4}\left(\frac{1}{z^{4}}\right)+4z^2\left(\frac{1}{z^{6}}\right)+\left(\frac{1}{z^2}\right)^4$
5

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$z^{8}+4z^{4}+6z^{4}\left(\frac{1}{z^{4}}\right)+4z^2\left(\frac{1}{z^{6}}\right)+\frac{1}{z^{8}}$

Final answer to the exercise

$z^{8}+4z^{4}+6z^{4}\left(\frac{1}{z^{4}}\right)+4z^2\left(\frac{1}{z^{6}}\right)+\frac{1}{z^{8}}$

Try other ways to solve this exercise

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  • Product of Binomials with Common Term
  • FOIL Method
  • Find the integral
  • Find the derivative
  • Factor
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
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Main Topic: Special Products

Special products is the multiplication of algebraic expressions that follow certain rules and patterns, so you can predict the result without necessarily doing the multiplication.

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