Exercise$\left(\frac{3}{7}x^5-\frac{1}{2}y^4\right)^4$
Step-by-step Solution
Intermediate steps
1
Expand the binomial $\left(\frac{3}{7}x^5-\frac{1}{2}y^4\right)^4$
$\left(\frac{3}{7}x^5\right)^4-2\left(\frac{3}{7}x^5\right)^3y^4+6\left(\frac{3}{7}x^5\right)^2\left(-\frac{1}{2}y^4\right)^2+\frac{12}{7}x^5\left(-\frac{1}{2}y^4\right)^3+\left(-\frac{1}{2}y^4\right)^4$
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Intermediate steps
2
The power of a product is equal to the product of it's factors raised to the same power
$\frac{81}{2401}x^{20}-2\cdot \left(\frac{27}{343}\right)x^{15}y^4+6\cdot \left(\frac{9}{49}\right)x^{10}\left(-\frac{1}{2}y^4\right)^2+\frac{12}{7}x^5\left(-\frac{1}{2}y^4\right)^3+\left(-\frac{1}{2}y^4\right)^4$
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Intermediate steps
3
Multiply the fraction and term in $-2\cdot \left(\frac{27}{343}\right)x^{15}y^4$
$\frac{81}{2401}x^{20}-\frac{54}{343}x^{15}y^4+6\cdot \left(\frac{9}{49}\right)x^{10}\left(-\frac{1}{2}y^4\right)^2+\frac{12}{7}x^5\left(-\frac{1}{2}y^4\right)^3+\left(-\frac{1}{2}y^4\right)^4$
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Intermediate steps
4
Multiply the fraction and term in $6\cdot \left(\frac{9}{49}\right)x^{10}\left(-\frac{1}{2}y^4\right)^2$
$\frac{81}{2401}x^{20}-\frac{54}{343}x^{15}y^4+\frac{54}{49}x^{10}\left(-\frac{1}{2}y^4\right)^2+\frac{12}{7}x^5\left(-\frac{1}{2}y^4\right)^3+\left(-\frac{1}{2}y^4\right)^4$
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Final answer to the exercise $\frac{81}{2401}x^{20}-\frac{54}{343}x^{15}y^4+\frac{54}{49}x^{10}\left(-\frac{1}{2}y^4\right)^2+\frac{12}{7}x^5\left(-\frac{1}{2}y^4\right)^3+\left(-\frac{1}{2}y^4\right)^4$