Question Video: Using Substitution of Terms to Verify an Identity Mathematics

Video Transcript

Let 𝑙 equal one-half times six to the 𝑥 power plus six to the negative 𝑥 power and 𝑚 equal one-half times six to the 𝑥 power minus six to the negative 𝑥 power. Is it true that 𝑙 squared minus 𝑚 squared equals one?

If we plug in what we know for 𝑙 and 𝑚, we get the expression one-half times six to the 𝑥 power plus six to the negative 𝑥 power squared minus one-half times six to the 𝑥 power minus six to the negative 𝑥 power squared. Because we’re going to have to be squaring some of these terms, I notice that we have six to the 𝑥 power and six to the negative 𝑥 power in both the 𝑙- and the 𝑚-variable.

One strategy to make our calculations a bit simpler would be to let 𝑎 equal six to the 𝑥 power and 𝑏 equal six to the negative 𝑥 power. If we do that, we end up with one-half times 𝑎 plus 𝑏 squared minus one-half times 𝑎 minus 𝑏 squared. We distribute our square to the one-half and the 𝑎 plus 𝑏 term and to the one-half and 𝑎 minus 𝑏 term. One-half squared is one-fourth. And we also know the form for 𝑎 plus 𝑏 squared. It’s equal to 𝑎 squared plus two 𝑎𝑏 plus 𝑏 squared. So we’ll add that. Bring down the subtraction. One-half squared is one-fourth.

And we have to be careful here because 𝑎 minus 𝑏 squared is not the difference of squares. 𝑎 minus 𝑏 squared will be equal to 𝑎 squared minus two 𝑎𝑏 plus 𝑏 squared. So we substitute 𝑎 squared minus two 𝑎𝑏 plus 𝑏 squared back into our equation. At this point, we have a one-fourth term that we could remove. We are undistributing that one-fourth. So we’ll have one-fourth multiplied by 𝑎 squared plus two 𝑎𝑏 plus 𝑏 squared. And then we’re subtracting this entire term, which means we’ll have negative 𝑎 squared. Subtracting negative two 𝑎𝑏 means we’ll be adding two 𝑎𝑏, and then subtracting a positive 𝑏 squared. So we have minus 𝑏 squared.

Before we do anything else, we wanna check and see if we can simplify by combining like terms. We have a positive 𝑎 squared term and a negative 𝑎 squared term, which cancels out. We have a positive 𝑏 squared term and a negative 𝑏 squared term, which also cancels out. And then we have two 𝑎𝑏 plus two 𝑎𝑏. We combine them by combining their coefficients so that we have four 𝑎𝑏. And if four 𝑎𝑏 is multiplied by one-fourth, one-fourth times four equals one. And we’re left with 𝑎 times 𝑏.

Now we need to go back in and plug in the values we know for 𝑎 and 𝑏. We’ll multiply six to the 𝑥 power times six to the negative 𝑥 power. To do that, we’ll have to remember our exponent rules when we have the same base taken to different powers being multiplied together. 𝑥 to the 𝑎 power times 𝑥 to the 𝑏 power is equal to 𝑥 to the 𝑎 plus 𝑏 power. Since both of these exponents have a base of six, we’ll combine their powers, which will be 𝑥 plus negative 𝑥. 𝑥 plus negative 𝑥 equals zero, and six to the zero power equals one.

Now, if we go back and look at our original question, we want to know is 𝑙 squared minus 𝑚 squared equal to one. We began with 𝑙 squared minus 𝑚 squared, which gave us 𝑎 times 𝑏. And then we found that 𝑎 times 𝑏 was equal to one. And so we can say, under these conditions, yes, 𝑙 squared minus 𝑚 squared is equal to one.