What is (3y^4+(2n)/7)*(3y^4-2x^2) ?

The solution to (3y^4+(2n)/7)*(3y^4-2x^2) is ((21y^4+2n)(3y^4-2x^2))/7

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simplify (3y^4+(2n)/7)*(3y^4-2x^2)

simplify

(3 y^{4} + \frac{2 n}{7}) \cdot (3 y^{4} - 2 x^{2})

\frac{( 21 y ^{4} + 2 n ) ( 3 y ^{4} - 2 x ^{2} )}{7}

Solution steps

(3 y^{4} + \frac{2 n}{7}) (3 y^{4} - 2 x^{2})

Join

3 y^{4} + \frac{2 n}{7} : \frac{21 y ^{4} + 2 n}{7}

= \frac{21 y ^{4} + 2 n}{7} (3 y^{4} - 2 x^{2})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 21 y ^{4} + 2 n ) ( 3 y ^{4} - 2 x ^{2} )}{7}

(- 1 + 3 i)^{25}

\frac{2}{( x ^{2} - 9 ) ^{2}}

(x + 1) ((x - 3) - 2) ((x - 3) + 2)

(- 2 x - 9 y = - 2, 5) (- 4 x - 9 y = - 23)

(2 - 3 sin (x))^{2}

What is (3y^4+(2n)/7)*(3y^4-2x^2) ?
The solution to (3y^4+(2n)/7)*(3y^4-2x^2) is ((21y^4+2n)(3y^4-2x^2))/7