What is (x+y)^2(x-y)^2(x^4+x^2y^2+y^4)^2 ?

The solution to (x+y)^2(x-y)^2(x^4+x^2y^2+y^4)^2 is (x+y)^2(x-y)^2(x^4+x^2y^2+y^4)^2

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simplify (x+y)^2(x-y)^2(x^4+x^2y^2+y^4)^2

simplify

(x + y)^{2} (x - y)^{2} (x^{4} + x^{2} y^{2} + y^{4})^{2}

(x + y)^{2} (x - y)^{2} (x^{4} + x^{2} y^{2} + y^{4})^{2}

Solution steps

(x + y)^{2} (x - y)^{2} (x^{4} + x^{2} y^{2} + y^{4})^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m},

assuming

a \geq 0, b \geq 0

(x + y)^{2} (x - y)^{2} (x^{4} + x^{2} y^{2} + y^{4})^{2} = ((x + y) (x - y) (x^{4} + x^{2} y^{2} + y^{4}))^{2}

= ((x + y) (x - y) (x^{4} + x^{2} y^{2} + y^{4}))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= (x + y)^{2} (x - y)^{2} (x^{4} + x^{2} y^{2} + y^{4})^{2}

\frac{1}{( 2 π )} \cdot e^{- (\frac{( 0.15 )}{2})^{2}}

e^{- 0.5256}

e^{(- 2 ∣ t ∣)} sin (3 t)

\frac{7 ^{\frac{1}{3}}}{4 ^{\frac{1}{3}}}

80 e^{- 0.1732800000000 \dots}

What is (x+y)^2(x-y)^2(x^4+x^2y^2+y^4)^2 ?
The solution to (x+y)^2(x-y)^2(x^4+x^2y^2+y^4)^2 is (x+y)^2(x-y)^2(x^4+x^2y^2+y^4)^2