To solve the equation (x/2)^3 = 64, rewrite it as x^3/8 = 64, then express 64 as 4^3 to get x^3 = 8^3. Using the difference of cubes identity a^3 - b^3 = (a-b)(a^2 + ab + b^2), factor to (x-8)(x^2 + 8x + 64) = 0. Applying the zero product rule gives x = 8 as the first solution. For the quadratic x^2 + 8x + 64 = 0, use the quadratic formula x = [-b ± √(b^2 - 4ac)]/(2a) with a=1, b=8, c=64, yielding x = -4 ± 4i√3. The three solutions are x = 8, x = -4 + 4i√3, and x = -4 - 4i√3.
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Olympiad Mathematics | Indian | can you solve this?Added:
Okay, if you're ready, let's solve this problem here.
This is some a quick one.
We have X over two all to the power of three equals 64.
So, how do we solve this problem here?
You know, we can always split what we have here. You see that the base you know, the power will be for both numerator and denominator.
So, this is the same as X to the power of three over two to the power of three equals 64.
Okay, since we are writing everything in the power of three let's write 64 in the power of three because 64 is a perfect cube.
Write X to the power of three over two to the power of three equals 64 is four to the power of three.
4 * 4 is 16, 16 * 4 and it's 64.
Right?
Now, cross multiply and we have X to power three to be equal to two to power three times four to power three.
And at this point, we shall apply a law of indices that says A to the power of M times B to the power of M is the same as A times B both to the power of M.
Right? So, if this is true, then we shall work on the right-hand side to get 2 * 4, that is eight. Then, it is to the same power of three.
Simple, right?
So, at this point, we have from two cubes we have X cubed equals eight cubed. So, ordinarily, you can agree with me that our x is equal to eight.
But, that will give us only one solution.
>> [snorts] >> So, we have to do this, bring this to the left, we have eight to the power of three, which is equal to zero at this point.
And then, we are going to understand that a to the power of three minus b to the power of three is an identity, which is equal to a minus b and we'll multiply a squared plus ab plus b squared.
Right? So, this is the identity for difference of two cubes.
Now, our a minus b shall be x minus eight.
Then, a squared shall be x squared.
ab is going to be um x times eight, and that is eight x.
Then, b squared is going to be eight squared.
Yes, b squared is going to be um eight squared.
And let me write it as eight squared.
If you want, you can write it as 64, but I prefer writing it as eight squared.
So, at this point, we apply our zero product rule.
A lot of students don't know when to apply this.
We apply zero product rule when you have two terms to multiply to get zero.
x minus one is considered a term.
Then, x squared plus eight x plus um eight squared is also a term. So, we're going to say that either this is equal to zero or this is equal to zero.
So, let's go.
Okay, so from here, applying the zero product rule, x - 8 is equal to zero. I'll pick this after.
And from here, our x is going to be zero plus eight. Meaning that the value of x is eight.
This is our first solution.
And we need to get two more solutions, which will come out of this.
We have x squared plus eight x plus eight squared equals zero.
I told you, if you want, you can write your you can write your eight squared as 64.
It doesn't change a thing.
Now, we are going to solve this quadratically using quadratic formula.
And that formula has a, which is the coefficient of x squared, and it's one.
It has b, which is the coefficient of x, and it's eight.
And then it has c, which is the constant, and in this case, it is eight squared.
So, what then is the formula?
The formula is x equals minus b plus or minus the square root of b squared minus four ac.
All of this is divided by two times a.
Okay, so once you're able to remember this formula, the work is almost done.
Just direct substitution.
Our x will be minus b, which will now be minus eight.
Right? Plus or minus, sorry, I'm writing out of sight. So, our b squared is eight squared now.
Minus four times one, cuz a is one, and c is what?
C is eight squared.
So, we divide all of this by two times one, which will be which will still be two.
Now, going on from here, we have x to be minus eight plus minus the square root of If you look at this term and this term, you know, four times one times eight squared is considered a term.
And what is common to them is eight squared. So, we write eight squared as the common factor.
Eight [snorts] squared divided by itself is one. Then, this the whole of this is going to be four times one, which is four.
Because eight squared here is already out.
We divide by two.
Now, to go on, we shall have x to be equal to minus eight plus minus We have the square root of eight squared times minus three.
Yes.
And we're dividing all of this by two.
And I hope you remember this um rule that if you have the square root of a times b, you can split it up and get the square root of a multiplied by the square root of b.
Yes, I said it. We can always do this.
So, from here now, guess what we are going to do.
We will now work on this like this.
So, permit me to remove this.
So, our x will be minus eight plus or minus we have the square root of eight squared multiplied by the square root of minus three.
And all of this is still over two.
Okay, so we can find the square root of this separately. Let's do it.
Okay, so from here now, we shall have x to be equal to minus eight plus or minus um this can take this out. so we have eight.
Then we have square root of three multiplied by square root of -1.
So, what I've done is to split this.
I will divide all of these by two.
Okay, I want to show all the steps.
Our X now is minus eight plus or minus where we already have eight, but square root of -1 is I.
So, eight times I will give us eight I.
Then we have our root three divide this by two.
So, that at the end of the day two here can divide the two numerators. So, that X will be minus eight divided by two, that is minus four.
Plus or minus eight I divided by two is four I.
Then this is root three.
Now we have a two-in-one solution.
Let's bring the three solutions together.
We got X to be equal to eight as our first solution.
And from this last part we have um we have minus four plus four I root three as our second solution.
Then the third solution still comes from the last part.
Where we have minus four minus four I.
Then we have root three. So, these three are the solutions to the equation.
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