What is (j^1+1+j^2)(j^1+1-j^2) ?

The solution to (j^1+1+j^2)(j^1+1-j^2) is (j+1+j^2)(j+1-j^2)

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simplify (j^1+1+j^2)(j^1+1-j^2)

simplify

(j^{1} + 1 + j^{2}) (j^{1} + 1 - j^{2})

(j + 1 + j^{2}) (j + 1 - j^{2})

Solution steps

(j^{1} + 1 + j^{2}) (j^{1} + 1 - j^{2})

Apply exponent rule:

a^{1} = a

j^{1} = j

= (j + 1 + j^{2}) (j + 1 - j^{2})

5 x - 3 y 8

4 x + 4 x + 3 x + 8 x + 8 x + 2 x

\frac{1}{2 ( e ^{3} ) ^{t} cos ^{2} ( t )}

x^{\frac{- 1}{3}} e^{x}

a \cdot (2 a + b)

What is (j^1+1+j^2)(j^1+1-j^2) ?
The solution to (j^1+1+j^2)(j^1+1-j^2) is (j+1+j^2)(j+1-j^2)