What is (1+it+t^2)^3 ?

The solution to (1+it+t^2)^3 is (t^6+1)+(3t^5+5t^3+3t)i

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simplify (1+it+t^2)^3

simplify

(1 + i t + t^{2})^{3}

(t^{6} + 1) + (3 t^{5} + 5 t^{3} + 3 t) i

Solution steps

(1 + i t + t^{2})^{3}

Apply exponent rule:

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

(1 + i t + t^{2})^{3} = (1 + i t + t^{2}) (1 + i t + t^{2})^{2}

= (1 + i t + t^{2}) (1 + i t + t^{2})^{2}

Apply exponent rule:

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

(1 + i t + t^{2})^{2} = (1 + i t + t^{2}) (1 + i t + t^{2})

= (1 + i t + t^{2}) (1 + i t + t^{2}) (1 + i t + t^{2})

Expand

(1 + i t + t^{2}) (1 + i t + t^{2}) : t^{4} + t^{2} + 2 t i + 2 t^{3} i + 1

= (t^{4} + t^{2} + 2 t i + 2 t^{3} i + 1) (1 + i t + t^{2})

Distribute parentheses

= t^{4} \cdot 1 + t^{4} i t + t^{4} t^{2} + t^{2} \cdot 1 + t^{2} i t + t^{2} t^{2} + 2 t i 1 + 2 t ii t + 2 t i t^{2} + 2 t^{3} i 1 + 2 t^{3} ii t + 2 t^{3} i t^{2} + 1 \cdot 1 + 1 i t + 1 \cdot t^{2}

Simplify

t^{4} \cdot 1 + t^{4} i t + t^{4} t^{2} + t^{2} \cdot 1 + t^{2} i t + t^{2} t^{2} + 2 t i 1 + 2 t ii t + 2 t i t^{2} + 2 t^{3} i 1 + 2 t^{3} ii t + 2 t^{3} i t^{2} + 1 \cdot 1 + 1 i t + 1 \cdot t^{2} : t^{6} + 3 t^{5} i + 5 t^{3} i + 3 t i + 1

= t^{6} + 3 t^{5} i + 5 t^{3} i + 3 t i + 1

Rewrite in standard complex form:

(t^{6} + 1) + (3 t^{5} + 5 t^{3} + 3 t) i

= (t^{6} + 1) + (3 t^{5} + 5 t^{3} + 3 t) i

\frac{5 x ^{2}}{( x ^{2} - 9 ) ^{2}}

4 (5 - sin (x))

7 (6 + 4)

(2 x y) (cos (t)) + (x^{2} - 2 y) (e^{t})

(x y - y) (x y + x)