What is (2i-2j+k)^2 ?

The solution to (2i-2j+k)^2 is (4j^2-4jk+k^2-4)+(-8j+4k)i

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simplify (2i-2j+k)^2

simplify

(2 i - 2 j + k)^{2}

(4 j^{2} - 4 jk + k^{2} - 4) + (- 8 j + 4 k) i

Solution steps

(2 i - 2 j + k)^{2}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

(2 i - 2 j + k)^{2} = (2 i - 2 j + k) (2 i - 2 j + k)

= (2 i - 2 j + k) (2 i - 2 j + k)

Distribute parentheses

= 2 i 2 i + 2 i (- 2 j) + 2 ik - 2 j \cdot 2 i - 2 j (- 2 j) - 2 jk + k \cdot 2 i + k (- 2 j) + kk

Simplify

2 i 2 i + 2 i (- 2 j) + 2 ik - 2 j \cdot 2 i - 2 j (- 2 j) - 2 jk + k \cdot 2 i + k (- 2 j) + kk : 4 j^{2} - 8 ji + 4 ki - 4 jk + k^{2} - 4

= 4 j^{2} - 8 ji + 4 ki - 4 jk + k^{2} - 4

Rewrite in standard complex form:

(4 j^{2} - 4 jk + k^{2} - 4) + (- 8 j + 4 k) i

= (4 j^{2} - 4 jk + k^{2} - 4) + (- 8 j + 4 k) i

(1 + cos (θ)) (1 - sec (θ))

\frac{( 3 x - 1 )}{x ( 3 x + 1 )}

\frac{x - 9}{( x - 1 ) ( x + 2 )}

2 (11 + 7)

\frac{3 !}{3 !}