What is expand (1+jsqrt(3))^6 ?

The solution to expand (1+jsqrt(3))^6 is 1+6sqrt(3)j+45j^2+60sqrt(3)j^3+135j^4+54sqrt(3)j^5+27j^6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (1+jsqrt(3))^6

expand

(1 + j 3)^{6}

1 + 63 j + 45 j^{2} + 603 j^{3} + 135 j^{4} + 543 j^{5} + 27 j^{6}

Solution steps

(1 + j 3)^{6}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 1, b = j 3

= i = 0 \sum 6 (i 6) \cdot 1^{(6 - i)} (j 3)^{i}

Expand summation

= \frac{6 !}{0 ! ( 6 - 0 ) !} \cdot 1^{6} (j 3)^{0} + \frac{6 !}{1 ! ( 6 - 1 ) !} \cdot 1^{5} (j 3)^{1} + \frac{6 !}{2 ! ( 6 - 2 ) !} \cdot 1^{4} (j 3)^{2} + \frac{6 !}{3 ! ( 6 - 3 ) !} \cdot 1^{3} (j 3)^{3} + \frac{6 !}{4 ! ( 6 - 4 ) !} \cdot 1^{2} (j 3)^{4} + \frac{6 !}{5 ! ( 6 - 5 ) !} \cdot 1^{1} (j 3)^{5} + \frac{6 !}{6 ! ( 6 - 6 ) !} \cdot 1^{0} (j 3)^{6}

Simplify

\frac{6 !}{0 ! ( 6 - 0 ) !} \cdot 1^{6} (j 3)^{0} : 1

Simplify

\frac{6 !}{1 ! ( 6 - 1 ) !} \cdot 1^{5} (j 3)^{1} : 63 j

Simplify

\frac{6 !}{2 ! ( 6 - 2 ) !} \cdot 1^{4} (j 3)^{2} : 45 j^{2}

Simplify

\frac{6 !}{3 ! ( 6 - 3 ) !} \cdot 1^{3} (j 3)^{3} : 603 j^{3}

Simplify

\frac{6 !}{4 ! ( 6 - 4 ) !} \cdot 1^{2} (j 3)^{4} : 135 j^{4}

Simplify

\frac{6 !}{5 ! ( 6 - 5 ) !} \cdot 1^{1} (j 3)^{5} : 543 j^{5}

Simplify

\frac{6 !}{6 ! ( 6 - 6 ) !} \cdot 1^{0} (j 3)^{6} : 27 j^{6}

= 1 + 63 j + 45 j^{2} + 603 j^{3} + 135 j^{4} + 543 j^{5} + 27 j^{6}

\frac{x ^{2} ( x - 4 )}{x + 5}

(1 + sin (u))^{2}

\frac{- 2 x - h}{( x - 1 ) ( x + h + 1 )}

10 (3 + 7)

960 (s + s^{'})^{6}

What is expand (1+jsqrt(3))^6 ?
The solution to expand (1+jsqrt(3))^6 is 1+6sqrt(3)j+45j^2+60sqrt(3)j^3+135j^4+54sqrt(3)j^5+27j^6