Use the Binomial Theorem to expand and then simplify the result: ...

Question

Use the Binomial Theorem to expand and then simplify the result: (x² +x+ 1)³.

Accepted Answer

Hey everyone here we are asked to expand the expression the quantity of two X squared plus X plus three raised to the third power using the binomial theorem. So before we can begin expanding, we need to recall the binomial theorem which gives us the formula for expanding binomial. So for any positive integer N we have the quantity of A plus B. To the end power is eight to the coefficient. We start with N zero times A. To the power of N plus N. One times A. To the power of n minus one times B plus n. Two times A. To the power of n minus two times B squared plus N. Three times A. To the power of n minus three times be cute. And then we continue expanding until we reach plus the next term. We have N N times B to the power of N. So with this formula in mind we can use it to expand our given expression. So we want to start by identifying the variables in the formula in relation to our given expression. So we can start with a. Well, A is going to be the left most term from our expression. So A will be two X squared. And this leaves us with B being the other two terms. So be will be X plus three. And lastly we need to find our value for N. Well, n this is just the power to which our expression is raised to. So our end is just going to be three. And now with this information identified. We can go ahead and start expanding. So again we have the expression the quantity of two X squared plus X plus three raised to the third power. And now expanding our first coefficient again we replace N with three. So we have 30 times A. Which is two X squared to the power of three. Now moving on, we add the next term. The coefficient is 31 times two X squared to the power of three minus one times B. Which again we identified B as X plus three. And now we can move on to our next term. We have +32 times the quantity of two X squared raised to the power of three minus two times the quantity of X plus three squared. And now moving on to our final term, we have the coefficient 3. 3 times the quantity of two X squared raised to the power of three minus three times B. Which is the quantity of X plus three. And we need to take the cube, raise this this expression to the power of three. And with this we are done expanding so we can go ahead and start simplifying. So again we have the expression the quantity of two X squared plus X plus three raised to the third power. And so with our first term here we can keep the coefficient the same for now we have +30 times now the quantity of two X squared to the power of three can be rewritten as eight X to the sixth power. Now moving on to our next term, we again keep the coefficient as 31. And then our quantity of two X squared is raised to the power of three minus one which is really raised to the second power. So we can rewrite this as four X to the fourth power. And then our next part of the expression just remains the same. So we have times the quantity of X plus three and then we can move on. So we have the coefficient of our next term which is times the quantity of two X squared raised to the power of three minus two is the same as raising it to the power of one. So it is just two X squared when we simplify and then four. Our next part of this term we have the quantity of X plus three squared and now squaring this expression, we end up with X squared plus six X plus nine and now we can move on to our final term. Again our coefficient here is +33 and our quantity of two X squared is raised to the power of three minus three which is just zero. So this term ends up just being one and then our quantity of X plus three is raised to the power of three. And cubing this expression we get X cubed plus nine X squared plus X plus 27. And so now we can just continue simplifying. So again we have the quantity of two X squared plus X plus three raised to the third power is expanding to now our first term. We can just keep it the same for now. So we have +30 times eight X to the power of six plus our next term. We have 31 as the coefficient. And now we can multiply our four X to the fourth power by the X plus three expression. So multiplying these terms here we have four X to the fifth power plus 12 X to the fourth power. Now we can move on to our next term. We have +32 as the coefficient and now we can just multiply our terms together here and we get the quantity of two X. To the fourth power plus 12 X cubed plus 18 X squared. And now we can move on to our last term. I'm just going to rewrite this because our one can be ignored. So we have 33 as the coefficient times the quantity of X cubed plus nine X squared plus 27 X plus 27. And so with this we now need to start solving for our coefficients here. And so if we recall this coefficient in the form N. R is equal to n factorial divided by the quantity of n minus R factorial times R factorial. So with this our upper value in our coefficient here in the parentheses the upper value is N. And the lower value is our so we can start writing out our coefficients using this formula. So for our first term we have three factorial divided by the quantity of three minus zero factorial times zero factorial. And then again this is the coefficient of eight X to the sixth power. And now for our next term the coefficient it turns out to three factorial divided by the quantity of three minus one factorial times one factorial. And again this is the coefficient of the quantity here of four X to the fifth power plus 12 X to the fourth power. Moving on to our next coefficient we again have three factorial divided by the quantity of three minus two factorial times two factorial. And this is the coefficient of the quantity of two X. To the fourth power plus 12 X cubed plus X squared. And lastly moving on to the last term, the coefficient here is three factorial divided by the quantity of 3 -3 factorial times three factorial. And again this is the coefficient of the expression X cubed plus nine X squared plus 27 X plus 27. And so from here we just need to start simplifying our coefficients. So beginning again with our first term. Well all of these these terms in the coefficient cancel out. So we are just left with eight x to the sixth power. And next our coefficient just simplifies to three. So we have three times the quantity of four X to the fifth Power plus 12 X to the fourth power for our next term. The coefficient also is just three. So we have three times the quantity of two X to the fourth power plus 12 X cubed plus 18 X squared. And for our final term the coefficient is just one just like our first term. So we are just left with this expression here of X cubed plus nine X squared plus 27 X plus 27. And now we can just start simplifying this expression here by first factoring in the coefficient. So we have eight X to the sixth Power plus 12 X to the fifth power plus 36 X to the fourth power plus six X to the fourth power plus 36 X cubed plus 54 X squared plus X cubed plus nine X squared plus 27 X plus 27. And now we just need to simplify these terms here. So some of the terms with the same power, we have this 36 X to the fourth power. And we also have six X to the fourth power. And then we also have this 36 X cubed as well as just this one X cubed. And lastly the last simplification. We can make here we have 54 X squared and nine X squared. So making these simplifications here, we are left with eight X. To the sixth power plus. 12 X to the fifth power plus. Now adding the power, the excess to the fourth power, we have 42 X. To the fourth power Plus and then adding the x cubes. We have 37 x. cute plus, adding up our X squares. We have 63 X squared. And then our last terms we just have plus 27 X plus 27. And with this we are done expanding and simplifying our given expression which again is the quantity of two X squared plus X plus three raised to the power of three. And so with this we are left with answer choice. A thanks for watching. I hope you found this video helpful and I'll see you again next time.

Use the Binomial Theorem to expand and then simplify the result: (x² +x+ 1)³.

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