This video demonstrates how to solve the equation (x/4)^3 = 27 by applying exponent rules, cross-multiplication, and the difference of cubes formula to factor the equation into (x-12)(x^2 + 12x + 144) = 0, yielding one real solution x = 12 and two complex solutions x = -6 ± 6√3i, which can be verified by substitution.
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Germany | Can You Solve This? | Math Olympiad追加:
Hello, you're welcome. How do I solve this nice algebra problem? Solution from here.
From here, we can write this as X over 4 all raised to power 3 equals to here, 27 that's 3 raised to power 3.
Then, from here this follows when we have A over B all raised to power N is the same thing as A raised to power N over B raised to power N.
Then, what we have here becomes X raised to power 3 over 4 raised to power 3 equals to 3 raised to power 3.
Then, we can cross multiply here. This over 1 here.
X raised to power 3 times 1 still X raised to power 3 equals to 3 raised to power 3 times 4 raised to power 3. We still have that 3 raised to power 3 times 4 raised to power 3.
Then, also this follows when we have A raised to power N times B raised to power N is the same thing as A times B all raised to power N.
Then, we can write what we have as X raised to power 3 equals to 3 times 4 all raised to power 3.
Which is the same thing as X raised to power 3 equals to 12 raised to power 3.
Then, we take the right hand side to the left hand side.
And we have X raised to power 3 minus 12 raised to power 3 equals to 0 here.
Then, this follows when we have A raised to power three minus b raised to power three.
This same thing as a minus b into bracket open bracket a squared plus ab plus b squared.
Then this also becomes x minus 12 into brackets and open brackets x squared plus 12 x plus 12 squared close brackets equals to zero here.
Then we have two possible cases here.
x minus 12 equals to zero or we have x squared plus 12 x plus 144 equals to zero here.
Solving on this side, we have x equals to 12.
Which is a real solution here.
And here, we have a quadratic equation.
Here, a equals to one, b equals to 12 and c equals to 1 144.
Applying the quadratic formula here, which is x equals to minus b plus or minus square root of b squared minus 4 a c over 2 a.
This becomes x equals to minus 12 plus or minus square root of 12 squared minus 4 times 1 times 144 over 2 times 1.
Then we have x equals to minus 12 plus or minus square root of 12 squared that's 144 minus 4 * 1 * 144. We can leave it as 4 * 144.
Then, over 2.
Then, this becomes x equals to minus 12 plus or minus square root of 144 is common here but 144 into bracket 1 minus 4 here over 2a.
Next step here, we have x equals to minus 12 plus or minus square root of 144 times 1 minus 4 minus 3 over 2.
Then, we can separate this one. We have square root of a times b which can be written as root a times root b.
This here, we have x equals to minus 12 plus or minus root 144 times root minus 3 over 2.
This we have x equals to minus 12 plus or minus root 144 as 12 and we have root minus 3 as root 3 times minus 1 over 2.
We can also separate this and we have x equals to minus 12 plus or minus 12 root 3 times root minus 1 over 2.
And here, root minus 1 is same thing as i.
It's a complex number here.
That is we have x equals to -12 plus or minus 12 root 3 i over two.
That is Here, we can write this x equals to -12 over two plus or minus 12 root 3 i over two.
Yeah, this becomes x equals to -12 over two, that's -6 plus or minus Also, here we have six root 3 i.
We have two complex solutions here. Therefore, altogether in this problem, we have three solutions here. One real solution here and two complex solutions here.
When we write it out, we have x1 equals to 12.
x2 equals to -6 + 6 root 3 i and x3 equals to -6 -6 root 3 i.
We have three solutions here in this problem. Let's check if this satisfies this given problem.
When we substitute the value of x here in this equation given x over 4 all raised to power 3 equals to 27.
When x is 12 now, we have 12 over 4 all raised to power 3 Is it equals to 27 here?
Yeah. 12 over 4, that's 3.
Then raised to power 3 is equals to 27 here.
And 3 raised to power 3, 3 multiplied itself three times, that's 27.
Equals to 27 here.
That is, left hand side now equals to right hand side.
Hence, x equals to 12 satisfies this given problem.
When we check the other two complex solutions, which we also satisfy what we have. And thank you for watching. Don't forget these steps.
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