Question
Question: How do you simplify \({{\left( y-4 \right)}^{3}}\)?...
How do you simplify (y−4)3?
Solution
We first find the simplification of the given polynomial (y−4)3 according to the identity (a−b)3=a3−3a2b+3ab2−b3. We need to simplify the cubic polynomial of the sum of two numbers. We already have the identity of (a−b)2=a2+b2−2ab. We then multiply the term (a−b) on both sides of the identity. We solve the multiplication to find the simplified form of (y−4)3 by replacing with a=y;b=4.
Complete answer:
We need to find the simplified form of (y−4)3. This is the cube of difference of two numbers. We know that (a−b)2=a2+b2−2ab.
We need to multiply the term (a−b) on both side of the identity (a−b)2=a2+b2−2ab.
On the left side of the equation, we get (a−b)2(a−b)=(a−b)3.
On the right side we have (a2+b2−2ab)(a−b). We use multiplication and get
⇒(a2+b2−2ab)(a−b)=a2.a+a.b2−2ab×a−a2.b−b2.b+2ab.b=a3+ab2−2a2b−a2b−b3+2ab2=a3−3a2b+3ab2−b3
We also can take another form where
(a−b)3=a3−3a2b+3ab2−b3=a3−b3−3ab(a−b).
Now we replace the values for a=y;b=4 in the equation (a−b)3=a3−3a2b+3ab2−b3.
(y−4)3=y3−3y2×4+3y×42−43=y3−12y2+48y−64
Therefore, the simplified form of (y−4)3 is y3−12y2+48y−64.
Now we verify the result with an arbitrary value of y=2.
We have (y−4)3=y3−12y2+48y−64.
The left-hand side of the equation gives (y−4)3=(2−4)3=(−2)3=−8.
The left-hand side of the equation gives
y3−12y2+48y−64=23−12×22+48×2−64=8−48+96−64=−8
Thus, verified the result of (y−4)3=y3−12y2+48y−64.
Note: We also can use the binomial theorem to find the general form and then put the value of 3.
We have (a−b)n=nC0anb0−nC1an−1b1+nC2an−2b2−....+(−1)rnCran−rbr+....+(−1)nnCna0bn. We need to find the cube of sum of two numbers. So, we put n=3.
(a+b)3=3C0a3b0−3C1a3−1b1+3C2a3−2b2−3C3a3−3b3=a3−3a2b+3ab2−b3.
In this way we also simplify the term of (a+b)3.