Aug 23, 2022

hey, i'm presh talwalkar, reminding you to care about your decisions, simplify and prove your

to the following expression, you can't use calculators or computers. I received many email requests for this type of problem. First, can you solve it? Try this problem, and when you're ready, continue watching the video to find a solution. So how can we simplify this expression? First I'll write a as the number under the first cube root and then b will do that. be a conjugate square root that is below 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 in the end we are looking for a real number now we will set this whole expression which is the cube root of a plus the cube root of b we will equal it to x now we will cube both sides and then expand now let's simplify this expression first we will put the squares inside these cube roots now we will group these cube roots together we can write a squared as a a and b squared as bb notice we have an a b term here and a ba term here and they will be equal to each other what is the product of the conjugate square roots? the cool thing is that it's going to be a difference of squares and this simplifies to be negative five cubed, we'll plug it in and now negative 5 cubed if we take the cube root of that, we end up with negative 5. 3 times minus 5 equals negative 15 so we have negative 15 and we can group the remaining terms which will be the cube root of a plus the cube root of b now we remember that x was defined to be exactly this quantity so we can plug that in now a plus b will equal just 16 because the square root terms will cancel out so we'll end up with a simple polynomial expression x cubed equals 16 minus 15x if there is a solution to this equation we are looking for a real value for be the

to our original expression , so we write x cubed plus 15x minus 16 equals 0.
How can we solve this equation well? You can usually try special values ​​and you'll quickly see that x equals 1 is a solution to this equation, so x minus 1 will be a factor of this polynomial, so we can factor this into the following product of x minus 1 and the quadratic x squared plus x plus 16. We can solve the quadratic using the quadratic formula, but we will learn that quadratic roots involve non-zero imaginary parts, so r strange roots cannot be the

to our original problem because we are looking for a real value of x this means the

is x equals one wow this is a very unexpected

to a simple algebra problem now you could Recheck this exercise by writing it in tungsten alpha.
Will it give the

of one? Surprisingly, you may find that it gives a complex