What is expand P_{15}(P_{6}+P_{7}x+x^2)^3 ?

The solution to expand P_{15}(P_{6}+P_{7}x+x^2)^3 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand P_{15}(P_{6}+P_{7}x+x^2)^3

expand

P_{15} (P_{6} + P_{7} x + x^{2})^{3}

P_{15} x^{6} + 3 P_{15} P_{7} x^{5} + 3 P_{15} P_{6} x^{4} + 3 P_{15} P_{7}^{2} x^{4} + P_{15} P_{7}^{3} x^{3} + 6 P_{15} P_{6} P_{7} x^{3} + 3 P_{15} P_{6}^{2} x^{2} + 3 P_{15} P_{6} P_{7}^{2} x^{2} + 3 P_{15} P_{6}^{2} P_{7} x + P_{15} P_{6}^{3}

Solution steps

P_{15} (P_{6} + P_{7} x + x^{2})^{3}

(P_{6} + P_{7} x + x^{2})^{3} = x^{6} + 3 P_{7} x^{5} + 3 P_{6} x^{4} + 3 P_{7}^{2} x^{4} + P_{7}^{3} x^{3} + 6 P_{6} P_{7} x^{3} + 3 P_{6}^{2} x^{2} + 3 P_{6} P_{7}^{2} x^{2} + 3 P_{6}^{2} P_{7} x + P_{6}^{3}

= P_{15} (x^{6} + 3 P_{7} x^{5} + 3 P_{6} x^{4} + 3 P_{7}^{2} x^{4} + P_{7}^{3} x^{3} + 6 P_{6} P_{7} x^{3} + 3 P_{6}^{2} x^{2} + 3 P_{6} P_{7}^{2} x^{2} + 3 P_{6}^{2} P_{7} x + P_{6}^{3})

Distribute parentheses

= P_{15} x^{6} + P_{15} \cdot 3 P_{7} x^{5} + P_{15} \cdot 3 P_{6} x^{4} + P_{15} \cdot 3 P_{7}^{2} x^{4} + P_{15} P_{7}^{3} x^{3} + P_{15} \cdot 6 P_{6} P_{7} x^{3} + P_{15} \cdot 3 P_{6}^{2} x^{2} + P_{15} \cdot 3 P_{6} P_{7}^{2} x^{2} + P_{15} \cdot 3 P_{6}^{2} P_{7} x + P_{15} P_{6}^{3}

= P_{15} x^{6} + 3 P_{15} P_{7} x^{5} + 3 P_{15} P_{6} x^{4} + 3 P_{15} P_{7}^{2} x^{4} + P_{15} P_{7}^{3} x^{3} + 6 P_{15} P_{6} P_{7} x^{3} + 3 P_{15} P_{6}^{2} x^{2} + 3 P_{15} P_{6} P_{7}^{2} x^{2} + 3 P_{15} P_{6}^{2} P_{7} x + P_{15} P_{6}^{3}

(tan (x) - 4) (tan (x) + 4) (tan (x) - 2) (tan (x) + 2)

5 x (x - 8 x)

\frac{2 ( - x ^{2} + 7 )}{( x ^{2} + 7 ) ^{2}}

(1 + ln (2))^{3}

x^{2} (cos^{2} (y) - sin^{2} (y))

What is expand P_{15}(P_{6}+P_{7}x+x^2)^3 ?
The solution to expand P_{15}(P_{6}+P_{7}x+x^2)^3 is