The portal has been deactivated. Please contact your portal admin.

Question Video: Simplifying an Expression Involving Exponents Mathematics • 9th Grade

Expand (βˆ’(π‘ŽΒ³ + (1/π‘ŽΒ²))¹⁰)^(1/5), where π‘Ž is a real constant.

04:17

Video Transcript

Expand negative one times π‘Ž cubed plus one over π‘Ž squared raised to the 10th power all raised to the power of one over five, where π‘Ž is a real constant.

In this question, we’re asked to expand an expression. And we’re told that the value of π‘Ž is a real constant. To do this, let’s start by considering what we need to expand. We have two exponents over parentheses. The inner exponent is an integer. So we can expand this over the parentheses by using the binomial formula. However, if we did this, we would then get a very complicated expression. Instead, let’s try and simplify this first by using the laws of exponents.

To determine which law of exponents we should use first, let’s start by considering the order of operations. In the order of operations, we start with the expressions inside the parentheses. This means we would start with adding π‘Ž cubed and one over π‘Ž squared. Next, we would move on to finding the exponent. We would multiply this by negative one. And finally, we would find the fifth root of all of this expression. Therefore, the final operation is raising this to the one-fifth power. And we’re raising everything inside the parentheses to this power. We can see that this is a product.

Therefore, we’re going to start by distributing this exponent of one-fifth over the parentheses. We’ll do this by using one of the laws of exponents. 𝑏 times 𝑐 all raised to the power of 𝑛 is equal to 𝑏 to the power of 𝑛 multiplied by 𝑐 to the power of 𝑛. Applying this, we get negative one to the power of one over five multiplied by π‘Ž cubed plus one over π‘Ž squared to the 10th power all raised to the power of one over five.

And now we can evaluate the first factor, negative one to the power of one over five. We recall that this means the fifth root of negative one. In other words, it’s the negative number we raise to the fifth power to get negative one, which we know is negative one. So the first factor evaluates to give us negative one.

In the second factor, we see we’re raising a base to an exponent and then raising this all to another exponent. And we can recall that we can simplify this by using one of our laws of exponents. If we raise 𝑏 to the power of 𝑛 and raise all of this to the power of π‘š, then we multiply the exponents. We get 𝑏 to the power of 𝑛 multiplied by π‘š. Applying this, we get a new exponent of 10 multiplied by one-fifth, which is of course equal to two.

Therefore, we’ve simplified this expression to give us negative one times π‘Ž cubed plus one over π‘Ž squared all squared. Now, all we need to do is expand the parentheses. We can do this by using the binomial formula or the FOIL method. We’ll use the FOIL method. We’ll start by multiplying the first terms of each binomial. This gives us π‘Ž cubed times π‘Ž cubed. And we can simplify this by using one of our laws of exponents. If we multiply two exponential expressions with the same base, we add their exponents. In other words, 𝑏 to the power of 𝑛 multiplied by 𝑏 to the power of π‘š is equal to 𝑏 to the power of 𝑛 plus π‘š. So π‘Ž cubed times π‘Ž cubed will be π‘Ž to the power of three plus three, which is π‘Ž to the sixth power. So our first term is π‘Ž to the sixth power.

Next, let’s multiply the outer terms together. That’s π‘Ž cubed multiplied by one over π‘Ž squared. And there’s two ways of evaluating this expression. We can either subtract the exponents or we can use our laws of exponents to rewrite one over π‘Ž squared as π‘Ž to the power of negative two. This then allows us to use our other law of exponents. We need to add the exponents together. We get π‘Ž to the power of three minus two, which is π‘Ž to the first power, or just π‘Ž. In either case, we get the next term is π‘Ž.

The next step of the FOIL method tells us to multiply the inner two terms together. This gives us one over π‘Ž squared multiplied by π‘Ž cubed. This is the same as the last expression. It’s also equal to π‘Ž. So the third term is π‘Ž.

Finally, we need to multiply the last two terms together. This is equal to one over π‘Ž squared multiplied by one over π‘Ž squared, which we can simplify by using the same law of exponents. It’s equal to one divided by π‘Ž to the power of two plus two, which is π‘Ž to the fourth power. Therefore, the final term in this expansion is one over π‘Ž to the fourth power.

Now, we just need to simplify this expression. First, π‘Ž plus π‘Ž is equal to two π‘Ž. Then, all we need to do is distribute negative one over our parentheses. We multiply all three terms by negative one, and this gives us our final answer. We were able to show that negative one times π‘Ž cubed plus one over π‘Ž squared raised to the 10th power all raised to the power of one over five is equal to negative π‘Ž to the sixth power minus two π‘Ž minus one over π‘Ž to the fourth power.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.