What is expand (e-fk)(e^2-efk+fk^2) ?

The solution to expand (e-fk)(e^2-efk+fk^2) is -k^3f^2+ek^2f^2+ek^2f-2e^2kf+e^3

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expand (e-fk)(e^2-efk+fk^2)

expand

(e - f k) (e^{2} - e f k + f k^{2})

- k^{3} f^{2} + e k^{2} f^{2} + e k^{2} f - 2 e^{2} k f + e^{3}

Solution steps

(e - f k) (e^{2} - e f k + f k^{2})

= (e - k f) (e^{2} - e k f + k^{2} f)

Distribute parentheses

= e e^{2} + e (- e f k) + e f k^{2} + (- f k) e^{2} + (- f k) (- e f k) + (- f k) f k^{2}

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= e^{2} e - ee k f + e k^{2} f - e^{2} k f + e kk ff - k^{2} k ff

Simplify

e^{2} e - ee k f + e k^{2} f - e^{2} k f + e kk ff - k^{2} k ff : - k^{3} f^{2} + e k^{2} f^{2} + e k^{2} f - 2 e^{2} k f + e^{3}

= - k^{3} f^{2} + e k^{2} f^{2} + e k^{2} f - 2 e^{2} k f + e^{3}

lo g_{10} (x) (x^{3} - 2)

[e^{- \frac{t}{2}} \cdot u \cdot (t - 2)]

2 (2 + x 10)

\frac{θ ^{4} - 4 θ ^{3} + 2 θ ^{2} - 3 θ + 1}{( θ ^{2} + 1 ) ^{3}}

\frac{1}{( a ^{2} + x ^{2} ) ( b ^{2} + x ^{2} )}

What is expand (e-fk)(e^2-efk+fk^2) ?
The solution to expand (e-fk)(e^2-efk+fk^2) is -k^3f^2+ek^2f^2+ek^2f-2e^2kf+e^3