Aug 23, 2022

# WOW! A Most Amazing Answer

Hey, this is Presh Talwalkar reminding you to simplify your choices and prove your

to the following expression. You may not use calculators or computers. I've received many email requests for this type of issue. I think Akshat in India because you suggested it to me Can you figure it out first try this problem and when you are ready watch the video to find a solution. How can we simplify this expression? I first write a as the number under the first cube root and then let b be a conjugate square root under the second cube root. Notice that the cube root of a is a real number and the cube root of b is a real number, so their sum will also be a real number, so we end up looking for a real number a plus the cube root of b.
We equate this to x. Now we roll both sides and then we expand. Now let's simplify this expression. We're going to put the squares in these cube roots first. Now let's group these cube roots together. We can then write a squared as a and b squared as bb. Notice that we have an a b term here and a ba term here and they are equal to each other which is the product of square root conjugates The nice thing is that it will be a difference of squares and this simplifies to to be minus five to the power of three.
We replace that with and now minus five to the third power, if we take the cube root of that we get minus 5. 3 times minus 5 equals minus 15. so then we have minus 15 and we can group the remaining terms that are the cube root of a plus the cube root of b will be. We now recall that x was defined to be exactly that quantity so we can substitute. Now a plus b equals 16 because the square root terms cancel, so we end up with a simple polynomial expression: x to the power of three equals 16 minus 15x if there is a solution in this equation for which we are looking for a real value is the

to our original expression, so we write x to the power of three plus 15x minus 16 equals 0.
How can we solve this equation well? Usually you can try specific values ​​and you will quickly see that x equals 1 is a solution to this equation. So x minus 1 is a factor of this polynomial. We can factor this into the following product of x minus 1 and squared x squared plus x plus 16. We can solve the square using the quadratic formula, but we will learn that the square roots both involve non-zero imaginary parts, so they're extraneous roots can't be the

to our original problem because we're looking for a real value of x Looking for. This means that the

x is one.
Double check this exercise by entering it into Wolfram Alpha. If she gives the

of one, you might be surprised to find that she gives a complex