# Question Video: Simplifying Expressions Involving Powers of Complex Numbers Mathematics • 12th Grade

Simplify (1 β π)Β³ β (1 + π) (1 β π) + (1 + π)Β³.

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### Video Transcript

Simplify one minus π cubed minus one plus π times one minus π plus one plus π cubed.

To simply this expression, weβll begin by distributing each set of parentheses. Letβs begin by distributing one minus π cubed. Now, we could write this as one minus π times one minus π times one minus π and then distribute the first pair of parentheses and then multiply all of that by one minus π.

The binomial theorem tells us that for positive integer values of π, π plus π to the πth power is the sum from π equals zero to π of π choose π times π to the πth power times π to the power of π minus π. So, in the distribution of one minus π cubed, the first term is one cubed. Then, the second term is three choose one times one squared times negative π to the power of one.

The third term is three choose two times one times negative π squared. And then, the third term is negative π cubed. Letβs simplify this a little. One cubed is one, and three choose one is three. So, our second term becomes three times one times negative π, which is negative three π. Our third term is three choose two, which is three times π squared. But of course, we know that π squared is negative one. So, this simply becomes negative three.

And our third term is negative π cubed. But since we can write π cubed as π squared times π, and we know π squared is negative one, this becomes negative negative one times π, which is simply π. And so, we simplify fully. And we find that one minus π cubed is negative two minus two π.

Weβll now repeat this process for our second term. Thatβs one plus π times one minus π. Now, we could distribute the parentheses using something like the FOIL method. Or we can recall that the product of a complex number and its conjugate is the sum of the squares of the real and imaginary parts of that complex number. And remember, the complex conjugate of one plus π is found by changing the sign of the imaginary part, so itβs one minus π. Its real part is one, and its imaginary part is the coefficient of π, so itβs also one. So, one plus π times one minus π is one squared plus one squared, which is simply two.

Letβs move on to the third pair of parentheses. Weβve got one plus π cubed. Once again, we use the binomial theorem. And we get one cubed plus three choose one times one squared times π plus three choose two times one times π squared plus π cubed. We simplify a little, and we get one plus three π plus three π squared plus π cubed. And then, we use again the fact that π squared is negative one. So, our third term becomes negative three. And our fourth term becomes negative π.

Simplifying this time, and we find that one plus π cubed is negative two plus two π. We can now replace each parentheses in our original expression with these. We get negative two minus two π minus two plus negative two plus two π. Then, we see that negative two π plus two π is zero. So, this becomes negative two minus two minus two, which is negative six. And so, when we simplify one minus π cubed minus one plus π times one minus π plus one plus π cubed, we get negative six.