What is expand intervals(x^2+x-2)^6 ?

The solution to expand intervals(x^2+x-2)^6 is

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expand intervals(x^2+x-2)^6

expand

in t er v a l s (x^{2} + x - 2)^{6}

e l in t r v a s x^{12} + 6 e l in t r v a s x^{11} + 3 e l in t r v a s x^{10} - 40 e l in t r v a s x^{9} - 45 e l in t r v a s x^{8} + 126 e l in t r v a s x^{7} + 141 e l in t r v a s x^{6} - 252 e l in t r v a s x^{5} - 180 e l in t r v a s x^{4} + 320 e l in t r v a s x^{3} + 48 e l in t r v a s x^{2} - 192 e l in t r v a s x + 64 e l in t r v a s

Solution steps

in t er v a l s (x^{2} + x - 2)^{6}

(x^{2} + x - 2)^{6} = x^{12} + 6 x^{11} + 3 x^{10} - 40 x^{9} - 45 x^{8} + 126 x^{7} + 141 x^{6} - 252 x^{5} - 180 x^{4} + 320 x^{3} + 48 x^{2} - 192 x + 64

= e l in t r v a s (x^{12} + 6 x^{11} + 3 x^{10} - 40 x^{9} - 45 x^{8} + 126 x^{7} + 141 x^{6} - 252 x^{5} - 180 x^{4} + 320 x^{3} + 48 x^{2} - 192 x + 64)

= e l in t r v a s (x^{12} + 6 x^{11} + 3 x^{10} - 40 x^{9} - 45 x^{8} + 126 x^{7} + 141 x^{6} - 252 x^{5} - 180 x^{4} + 320 x^{3} + 48 x^{2} - 192 x + 64)

Distribute parentheses

= in t er v a l s x^{12} + in t er v a l s \cdot 6 x^{11} + in t er v a l s \cdot 3 x^{10} + in t er v a l s (- 40 x^{9}) + in t er v a l s (- 45 x^{8}) + in t er v a l s \cdot 126 x^{7} + in t er v a l s \cdot 141 x^{6} + in t er v a l s (- 252 x^{5}) + in t er v a l s (- 180 x^{4}) + in t er v a l s \cdot 320 x^{3} + in t er v a l s \cdot 48 x^{2} + in t er v a l s (- 192 x) + in t er v a l s \cdot 64

Apply minus-plus rules

+ (- a) = - a

= e l in t r v a s x^{12} + 6 e l in t r v a s x^{11} + 3 e l in t r v a s x^{10} - 40 e l in t r v a s x^{9} - 45 e l in t r v a s x^{8} + 126 e l in t r v a s x^{7} + 141 e l in t r v a s x^{6} - 252 e l in t r v a s x^{5} - 180 e l in t r v a s x^{4} + 320 e l in t r v a s x^{3} + 48 e l in t r v a s x^{2} - 192 e l in t r v a s x + 64 e l in t r v a s

\frac{( x ^{3} + y ^{3} + z ^{3} ) ( x y + x z + yz )}{x yz ( x ^{2} + y ^{2} + z ^{2} )}

\frac{1}{( - 1 - s ) ^{2}}

\frac{1}{2} (- \frac{ln ( x )}{5 x ^{5}} - \frac{1}{25 x ^{5}})

3 x + 5 x + 14 x

(x - 1) (x + 2) (x - 10) (x + 10)

What is expand intervals(x^2+x-2)^6 ?
The solution to expand intervals(x^2+x-2)^6 is