To solve the equation -K³/8 = 1, we first rearrange it to K³ + 8 = 0, which can be factored using the sum of cubes formula A³ + B³ = (A + B)(A² - AB + B²) as (K + 2)(K² - 2K + 4) = 0. This yields one real solution K = -2 and two complex solutions K = 1 ± i√3, which can be verified by substituting back into the original equation.
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Let's consider this nice algebraic equation here to find the value of K.
Solution from here, we have - K cubed over 8 equals to 1.
From here, this is also 1 over 1.
And when we cross multiply, K - K cubed times 1 1 * 8 becomes - K cubed equals to 8.
Then we can bring everything to one side, of course. - K cubed going here becomes plus, so we have 8 plus K cubed then equals to 0.
Which of course we can also write as K cubed plus 8 equals to 0.
Then let's express it in power of 3 as well, so we can say K cubed plus 2 cubed equals to 0.
Which implies this sum of two cubed follows where we have A cubed plus B cubed which can be expressed as A plus B into brackets, open brackets, A squared plus AB plus B squared.
That is what we have here becomes K plus 2 into brackets, open brackets, A squared.
Sorry, this is minus AB, so this minus 2K plus K squared close bracket equals to 0 here.
Then we have two possible cases, K plus 2 equals to 0 or we have K squared - 2K plus K squared equals to zero.
That is, this becomes k equals to minus two.
We have a real solution here. Here we have a quadratic equation. Here a equals to one, b equals to minus two, and c equals to I'm sorry, there is a mistake.
It should be b squared, and b is two here.
Uh b is two, so that's two squared.
Which is two squared becomes four. So, our c equals to four. All right. From here, we apply the quadratic formula. k equals to minus b plus or minus square root of b squared minus four ac all over two a.
That is, this becomes k equals to minus minus two plus or minus square root of minus two squared minus four times one times four all over two times one.
That is, then next step we have k equals to minus as minus plus that's two plus or minus square root of minus two squared you have four minus four times one times four it carries us four times four all over two.
That is, we have k equals to two plus or minus square root of Four is common here, we have four which will give us one minus four all over two.
We have k equals to two plus or minus square root of four times one minus four minus three all over 2 which same thing as K equals to 2 plus or minus square root of 4 * 3 * -1 over 2 We can separate this as K equals to 2 plus or minus root 4 * root 3 * root -1 over 2 But this way of K equals to 2 plus or minus root 4 that's 2 and we have root 3 root -1 is I all over 2 That is We can separate this as K equals to 2 over 2 plus or minus 2 root 3 I over 2 That is we have K equals to 2 over 2 that's 1 plus or minus 2 cancels each other here as well we have roots 3 I We have two complex solutions here.
Therefore Altogether we have three solutions here one real and two complex. Like this we can write K1 equals to -2 K2 equals to 1 plus root 3 I Then K3 equals to 1 minus root 3 I We have one real and two complex solution here.
Then when we check on what is given and substitute the value of K That is from what we have -K³ over 8 equals to 1 When we substitute K equals to minus two, we have minus into minus two cubed over eight, this is equals to one.
Then from here, we have minus times minus raised to power three, which still give us minus. So, we have minus two cubed as minus eight over eight, which is equals to one here.
Then minus times minus is plus. We have eight over eight, which is equals to one here.
And eight over eight is one, which is equals to one on this side.
Like this, we have the left hand side equals to the right hand side.
And therefore, we conclude that the value of k here, which is minus two, satisfy this given problem.
When we check for the other two complex solutions, it also satisfy this given problem.
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