This tutorial provides a clear and efficient roadmap for mastering algebraic expansions through systematic pattern-matching. However, it leans heavily on rote memorization, sacrificing conceptual depth for mechanical accuracy.
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Special Products: Cube of Binomials Grade 8 Math
Added:In this video, you are going to learn how to cube a binomial.
Suppose you have the cube of X + Y.
This is equivalent to X + Y times X + Y times X + Y. If you want to save up your time and minimize the error in doing algebra, you can use the pattern X cubed plus thrice the product of first term and then the second term.
But, you have to square the first term.
Okay?
Plus again, thrice the product of um first and then the second term, but you have to square the second term.
And lastly, you're going to cube the second term.
This is our first situation.
And lastly, we have the second situation.
The cube of X - Y in X - Y times X - Y times X - Y.
Now, this is X cubed 3 X squared Y plus 3 X Y squared Y cubed.
What happens here is that um the sign alternates. So, from the first term, we have positive. Second term, we have negative.
Positive, negative. If you want to um continue with this process, um you will end up with this one.
Okay, suppose we have examples.
Example number one.
Let's say the cube of 2x + y.
So, copying the pattern, we have cube the first term.
The first term we have 2x cubed + 3 * the square of the first term * the second term which is y + 3 * the first term * the square of the second term. And lastly, um the cube of the second term.
Observe if this is addition, we're using an operation addition.
Um all sign must be positive.
Now, um simplify the result. Uh the cube of 2x This will give you 2 cubed. That is 2 * 2 * 2 which is 4 * 2 which gives you 8.
So, you have 8 and then x cubed.
Right? And then plus you simplify this term.
That would be uh first you're going to do is you're going to bring down three.
You simplify 2x squared. That would give you 4 x squared * y.
And then plus 3 * 2x that would give you 6x and then y squared is just y squared. + y cubed.
All right, next.
You bring down 8x cubed plus what is 3 * 4x squared? That would definitely give give you 12x squared and then you bring down y plus 6x you bring down 6xy squared plus y cubed >> [clears throat] >> All right, um this would be the final answer.
As you can observe, there are no like terms. So, meaning to say this is your final answer.
A second example the cube of 3a minus b Okay.
Now um with the use of our guide earlier, we will be able to solve this more quickly.
>> [clears throat] >> So first is you're going to observe that we are now using an operation subtraction.
Meaning to say the sign alternates.
Going to cube the first term.
Okay.
Minus thrice the product of the first term and the second second terms 3a squared times b and then plus 3 times the product of 3a and b but you have to a square that one and lastly minus the cube of the second term. Now, simplifying we have 3 a cubed then that would mean 3 cubed 3 * 3 * 3, that would give you 27.
27 a cubed and then minus um this becomes 9.
3 squared is 9.
And then squared a squared times b plus 3 * 3, that would give you 9.
And then you just copy a and then b squared is just b squared.
minus b cubed Okay?
Now, we're not yet done. You bring down the first term and then what is um -3 * 9? That would definitely give you -27 a a squared b.
+9 a b squared minus b cubed This would be our final final answer.
Third example.
Suppose we have the cube of x squared plus y squared.
Repeating the pattern, we have cube the first term Okay?
plus thrice the product of the first term and the second term x squared times um y squared, but you have to square the first term plus thrice the product again of the first term and then the second term but you have to square the second term and lastly you are going to cube the second term.
Second is y squared.
All right.
We simplify the result. x squared raised to the power of three, that would give you x to the power of six.
Plus um three times 2 * 2, that would give you four.
And then you'll bring down y squared.
Plus three you bring down x squared.
And then y squared to the power of two, that would give you y to the power of four.
Plus y to the power of six.
Okay? So, as simple as that.
That's how you're going to cube a binomial.
Okay, one more example. Let's say we have number four.
Five, the cube of 5x squared plus 2y and then repeating the pattern we have um cube the first term.
5x squared cubed plus three times the product of the first term and then the second term.
And then 2y, but you have don't forget to uh square the first term.
Plus three times 5x squared times 2y squared plus the um cube of the second term.
We have 2y cubed.
All right.
Okay.
>> [clears throat] >> Now, what happens here is that you're going to cube the five or the numerical coefficient. That would be five cubed.
5 * 5 * 5 25 * 5 that would give you 125.
Okay?
So, that is 125 and then applying the power rule um x to the power of six. Simply multiply the exponents.
Plus um simplify the power rule.
Um we have three.
Five squared that would give you 25 x to the power of four.
And then you copy 2y.
Plus 3 * 5 that would give you 15 x squared.
And the square of 2y that would give you um 4y squared.
Plus 2 * 2 * 2 that would give you eight y cubed.
Simplifying further, um bring down 125 x to the power of six.
What is 25 * 3? That would definitely give you um 75 x to the power of four.
25 * 3 5 6 75.
>> [clears throat] >> And you have to multiply it back to 2y.
Plus 15 * 4 that would give you um 30 60.
60 x squared y squared and then plus eight y cubed.
We're not yet done.
Okay, let's continue.
Um bring down 125 x to the power of 6 plus 75. Okay, here.
Now, what is 75 * 2? 150 x to the power of 4 y plus 60 x squared y squared and plus 8y cubed. That would be the end of my presentation. I am hoping that you are able to get the idea on solving cube of binomials. So, first thing, you must familiarize the pattern.
>> [snorts] >> First situation is um addition and then the second situation is subtraction. For subtraction, um the sign alternates, positive, negative, positive, negative.
And for the first situation, as you can observe that it's always positive throughout the end of the terms.
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