Formulas for abbreviated multiplication. Formula for the cube of a binomial - $(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$

We continue with the next of the abbreviated multiplication formulas $(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$. Let's look at some problems to illustrate some of its applications.

Problem 1 Perform the grading $(x+2)^3$.

Solution: To solve this problem we will apply the formula $(a+b)^3=a^3+3a^2b+3ab^2+b^3$, where in our case $a=x$ and $b=2$, so we get $(x+2)^3=x^3+3x^2.2+3x.2^2+2^3=x^3+6x^2+12x+8$.

Problem 2 Perform the $(2m-3n)^3$ grading.

Solution: $(2m)^2.3n+3.(2m).(3n)^2-(3n)^3=8m^3-36m^2n+54mn^2-27n^3.$

3 Problem Perform the grading $(3t+z^2)^3$

Solution: To solve this problem we will apply the formula $(a+b)^3=a^3+3a^2b+3ab^2+b^3$, where $a=3t$ and $b=z^2$, hence $(3t+z^2)^3=(3t)^3+3. (3t)^2.z^2+3.(3t)(z^2)^2+(z^2)^3=27t^3+27t^2z^2+9z^4t+z^6.$

4 Problem Simplify the expression $(x+1)^3-2(x-1)^2.$

Solution: Notice that the expression we need to simplify involves two of the shortcut multiplication formulas $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ and $(a-b)^2=a^2-2ab+b^2$, the latter of which we have discussed in detail here. We apply them and obtain $(x+1)^3-2(x-1)^2=x^3+3x^2+3x+1-2(x^2-2x+1)=x^3+3x^2+3x+1-2x^2+4x-2=x^3+x^2+7x-1.$ 

5 Problem Simplify the expression $(3x+2)^3-3x(3x+1)(3x-1)-(3x+2)(x+4)$

Solution: We apply the formulas $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ and $(a-b)(a+b)=a^2-b^2$, which we have already covered here, and multiply the product $(3x+2)(x+4)$ by the "any by any" rule, следователно $(3x+2)^3-3x(3x+1)(3x-1)-(3x+2)(x+4)=9x^3+54x^2+32x+8-3x(9x^2-1)-(3x^2+12x+2x+8)=9x^3+54x^2+32x+8-27x^3+3x-3x^2-14x-8=-18x^3+51x^2+21x.$ 

6 Problem Perform the indicated operations $(a-1)^3-(a+1)^3-(a+1)(a-1).$

Solution: We apply the formulas $(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$ and $(a-b)(a+b)$ and obtain that $(a-1)^3-(a+1)^3-(a+1)(a-1)=a^3-3a^2+3a-1-(a^3+3a^2+3a+1)-(a^2-1)=a^3-3a^2+3a-1-a^3-3a^2-3a-1-a^2+1=-7a^2-1.$

7 Problem Represent a quadratic as the cube of a binomial $x^3-12x^2+48x-64.$

Solution: Notice in the given polynomial that $12x^2=3.4.x^2$, $48x=3.16.x$ and $64=4^3$, therefore we can write this polynomial in the form $x^3-3.x^2.2^2+3.x.4 ^2-4^3$, now when we substitute in the right hand side of the formula $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ $a=x$ and $b=4$, we get that $x^3-3.x^2.2^2+3.x.4^2-4^3=(x-4)^3.$ This is how we represented the given quadratic as a cube of a binomial.  

8 Problem Determine the value of the expression $(x+1)^3-(x-1)(x+1)-x(x+1)^2$, given $x=\frac{1}{2}.$

Solution: Apply the formulas $(a+b)^3=a^3+3a^2b+3ab^2+b^3$, $(a-b)(a+b)=a^2-b^2$, and $(a+b)^2=a^2+2ab+b^2$, therefore $(x+1)^3-(x-1)(x+1)-x(x+1)^2=x^3+3x^2+3x+1-(x^2-1)-x(x^2+2x+1)=x^3+3x^2+3x+1-x^2+1-x^3-2x^2-x=2x+2.$ Now we calculate the value of the expression for $x=\frac{1}{2}$ and get $2.\frac{1}{2}+2=3.$

Homework assignments

1. Perform the grading:

a) $(2x+1)^3;$ b) $(2m-2n)^3;$ c) $(\frac{1}{3}a+\frac{1}{3}b)^3;$ d) $(3-y^2)^3;$ e) $(a-b+c)^3.$

2. Simplify the expression: 

а) $(2x+5)^3-2x(2x-3)^2-5(2x+25);$ 

б) $(y+2)^3+(2y+x-3)-(-x+3)^2+(-y-2)^3;$ 

c) $t(t-3)^2-(t-1)^3;$ 

3. Prove that the value of the expression $(x+2)^3-6(x+1)(x-1)-3(4x+1)-x^3$ is independent of the value of $x.$

4. Find the numerical value of the expression: 

a) $(2x+1)^3-8x(x+1)(x-1)-6x(2x-3)$, with $x=(-\frac{1}{2})^5;$

b) $x(x-2)-(x-2)^3$ at $x=-1;$ c) $6t^2+(t-2)^3-t^3$ at $t=\frac{2}{3};$

d) $b(b-1)(1+b)+(-b-1)^3+b$ at $b=-^frac{1}{3};$ e) $z(z-1)(1+z)+(-z-1)^3$ at $z=(-1)^{303}.$

5. For which value of the parameter $c$ the coefficient in front of $x^2$ in the normal form of the polynomial that is equal to the expression $A=(c-2x)^3-x(x-c)$ is equal to 22.

6. For which value of the variable is the numerical value of the expression:

a) $(4-y)y(4+y)+(y-2)^3+6y^2$ is equal to $20;$ b) $x(x-1)^2-(x-1)^3-x^2$ is equal to -3;

7. Find the smallest value of the expression $(2+x)^3-3x(\frac{x^2}{3}+4)+(3-x)(3+x).$

8. Find the largest value of the expression $(y-3)^3-3y(9+\frac{y^2}{3})-(y+6)(y-6)$.

9. Find the normal polynomial identical to the expression: 

а) $(a^2-3)^3-(a-2)(a^2+4)(a+2)-a^4(a^2-10)$;

b) $(2a-3b)^3-(a+2b)^3+7a^2(6b-a)$;

c) $(2a-3)^3-(2-3a)^3-35a^3+90a^2$;

d) $(\frac{1}{2}x+2y)^3-(2x-\frac{1}{2}y)^3+7\frac{7}{8}x^3-7\frac{1}{2}x^2y$;

e) $x^3+3x^2(x-2)+3x(x-2)^2+(x-2)^3$;

f) $(a-b)^3+3(a-b)^2(b-c)+3(a-b)(b-c)^2+(b-c)^3$;

g) $(2x-m)^3-3(2x-m)^2(y-m)+3(2x-m)(y-m)^2-(y-m)^3$.