Perform the indicated operations and/or simplify each expression....

Question

Perform the indicated operations and/or simplify each expression. Assume all variables represent positive real numbers. ( 5/∛2) - ( 2/∛16) + (1/∛54)

Accepted Answer

Hello, today we're going to be simplifying the given expression. So what we are given is nine over the cube root of three minus seven over the cube root of 375 plus five over the cube root of 81. Now, before we combine these values together, let's go ahead and simplify the given denominators. The Q Bert of three is in its simplest form. So we're not going to be simplifying it any further. The Cube root of 30 375 while is not a perfect cube, but we can rewrite it as a product of prime numbers. So if we break up 375 by using a number tree, 375 can be rewritten as five times 75 where five is a prime number, 75 can be broken down as five, times 3 with three being a prime number and 25 can be broken down as five times five, which are both prime numbers. So the cube root of 375 can be rewritten as the cube root of three times five times itself three times which is five cubed. And the cube root allows us to take out any numbers that are in multiples of three, which in this case, five cubed is a multiple of three. So we can take out five cubed as a single five from the cube root. And this can be rewritten as five times the cube root of three. So that's going to be the simplified denominator for seven over the cube root of 75. Now let's simplify the denominator of the third term. Now, again, 81 is not a perfect cube, but we can rewrite it as a product of prime numbers. 81 can be rewritten as nine times nine and both nines can be rewritten as three times itself, which are both prime numbers. So what we have is the cube root of 81 and that can be rewritten as three times itself four times or in this case, we can rewrite it as three times three cubed. Now again, the Cube root allows us to take out any numbers in multiples of three, which in this case, three times itself, three times can be taken out as a single three. So the cube root of three times three cubed can be rewritten as three times the cube root of three. So this is going to be the simplified denominator of the third term. And if we replace the denominators, the new expression that we have is nine over the cube root of three -7/5 times the q bert of three Plus five over the cube root Or three times the cube root of three. Okay. So now if we want to add all these fractions together, we need to go ahead and get a common denominator between all three fractions. So in this case here, we need to look for the lowest common denominator. Well, every denominator here has the cube root of three. So the L C D is going to need to have the cube root of three. But now we need to figure out the L C D between the coefficients. In the first denominator, we have the coefficient of one. In the second denominator, we have the coefficient of five. And in the third denominator, we have the coefficient of three. So what's gonna be the lowest common denominator between 15 and three? While the lowest common multiple between three numbers is going to be 15. So we need to multiply each fraction by a value that's going to allow us to get 15 in the denominator. So in the first fraction, we can multiply nine over the cube root of three x 15/15. In the second fraction, we can multiply this by 3/ Because five times 3 is going to give us 15. And in the third fraction, we can multiply the last term by 5/5 because five times three is going to give us 15. So in the first denominator or in the first term, we have 15/15 times nine over the cube root of three, which is going to give us 135 over 15 times the cube root of three. The second term is going to simplify to give us over 15 times the cube root of three because seven times three is going to give us 21 3 times five cubit of three is going to give us 15 cube root of three. And the last term, five times five is going to give us 25 and five times Q three, Q bit of three is going to give us 15 cube root of three. Now notice that we have the same denominator in each term. Since we have the same denominator, we can go ahead and add up the numerator is together. So we have 100 and 35 minus 21 plus 25 which is going to give us over 15 cube root of three. Now, we want to go ahead and rationalize the denominator of this value because we don't want to have any cube roots in the denominator. In order to rationalize this, let's go ahead and multiply the numerator and denominator by the cube root of three squared. The reason why we want to multiply this by the cubit of three squared is because the cube root of three times the cube root of three squared can simplify to give us the cube root of three cubed. And just like before the cube root allows you to take out any numbers in groups of three, which in this case here, this is going to simplify to just give us three. So what we are left with is 100 and 39 times the cube root of three squared, which is going to be 100 and 39 times the cube root of three squared Over 15 cubit of three times the cube root of three squared, which is going to simplify to give us 15 times three. Now, in the numerator three squared is going to simplify to give us nine. So we have 139 times the cube root of nine. And in the denominator 15 times three is going to give us 45. And this is going to be the simplified form of the given expression. And with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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Perform the indicated operations and/or simplify each expression. Assume all variables represent positive real numbers. ( 5/∛2) - ( 2/∛16) + (1/∛54)

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