Video: Finding the Value of an Unknown by Factorizing the Sum of Two Cubes

If π‘₯Β³ + π‘˜Β³ = (π‘₯ + π‘˜)(π‘₯Β² βˆ’ 2π‘₯ + π‘˜Β²), what is the value of π‘˜?

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Video Transcript

If π‘₯ cubed plus π‘˜ cubed is equal to π‘₯ plus π‘˜ multiplied by π‘₯ squared minus two π‘₯ plus π‘˜ squared, what is the value of π‘˜?

There is a shortcut to answer this question, if you know how the sum of two cubes factorise. If not, the easiest method is to expand the brackets or parentheses, and then compare coefficients. We need to expand, or multiply out, π‘₯ plus π‘˜ and π‘₯ squared minus two π‘₯ plus π‘˜ squared.

Firstly, we’ll multiply each of the terms in the second parenthesis by π‘₯. π‘₯ multiplied by π‘₯ squared is equal to π‘₯ cubed. π‘₯ multiplied by negative two π‘₯ is equal to negative two π‘₯ squared. And finally, π‘₯ multiplied by π‘˜ squared is equal to π‘˜ squared π‘₯.

Our next step is to multiply all three terms in the second parenthesis by π‘˜. π‘˜ multiplied by π‘₯ squared is π‘˜π‘₯ squared. π‘˜ multiplied by negative two π‘₯ is negative two π‘˜π‘₯. And finally, π‘˜ multiplied by π‘˜ squared is equal to π‘˜ cubed. We can then group, or collect, the π‘₯ squared terms and, separately, the π‘₯ terms.

Grouping the π‘₯ squared terms gives us π‘˜ minus two π‘₯ squared, as we have π‘˜π‘₯ squared minus two π‘₯ squared. Grouping the π‘₯ terms gives us π‘˜ squared minus two π‘˜π‘₯. As we also have an π‘₯ cubed and a π‘˜ cubed term, we can say that π‘₯ plus π‘˜ multiplied by π‘₯ squared minus two π‘₯ plus π‘˜ squared is equal to π‘₯ cubed plus π‘˜ minus two π‘₯ squared plus π‘˜ squared minus two π‘˜π‘₯ plus π‘˜ cubed.

We know that this is equal to π‘₯ cubed plus π‘˜ cubed. As there is no π‘₯ squared or π‘₯ term in this expression, we know the following two facts. We know that π‘˜ minus two is equal to zero. And π‘˜ squared minus two π‘˜ is also equal to zero. Adding two to both sides of this first equation gives us π‘˜ is equal to two.

We can check this answer by substituting π‘˜ equals two into the second equation, giving us two squared minus two multiplied by two. Two squared is equal to four. And negative two multiplied by two is equal to negative four. Therefore, the answer of this is zero. We have, therefore, concluded that if π‘₯ cubed plus π‘˜ cubed is equal to π‘₯ plus π‘˜ multiplied by π‘₯ squared minus two π‘₯ plus π‘˜ squared, then π‘˜ is equal to two.

We mentioned at the start that the sum of two cubes, in this case π‘₯ cubed plus π‘˜ cubed, has a standard factorisation. This is equal to π‘₯ plus π‘˜ multiplied by π‘₯ squared minus π‘˜π‘₯ plus π‘˜ squared. As we can see from the initial equation, negative π‘˜ has been replaced by negative two. We can, therefore, say that negative π‘˜ is equal to negative two. And, once again, π‘˜ is equal to two.

This is a standard factorisation that works any time we’re dealing with the sum of two cubes.

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