Difference of squares: More than one factorization

Let's solve the equation. First, we'll simplify the algebraic expressions using the difference of squares formula:

$(a+b)(a-b)=a^2-b^2$We'll apply this formula and expand the parentheses in the expressions in the equation:

$(x+3)(x-3)=x^2+x \\ x^2-3^2=x^2+x \\ x^2-9=x^2+x$We'll continue and combine like terms. After moving terms around, we can see that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2-9=x^2+x \\ -9=x\\ \downarrow\\ \boxed{x=-9}$Therefore, the correct answer is answer B.

Solve the equation by simplifying the expression on the left side in two steps. First, we'll proceed to multiply the expressions in the two leftmost pairs of parentheses:

Apply the shortened multiplication formula for squaring a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

Due to the fact that these two pairs of parentheses are being multiplied by another expression (which is also in parentheses), place the result inside of parentheses (marked with an underline):

$\underline{ (x-1)(x+1)}(x-2)=-2x^2+x^3 \\ \underline{ (x^2-1^2)}(x-2)=-2x^2+x^3 \\ (x^2-1)(x-2)=-2x^2+x^3$

Continue to simplify the expression on the left side by using the expanded distribution law:

$(a+b)(c+d)=ac+ad+bc+bd$

Additionally, apply the law of exponents for multiplying terms with equal bases:

$a^ma^n=a^{m+n}$ We'll now apply these laws and expand the parentheses in the expression in the equation:

$(x^2-1)(x-2)=-2x^2+x^3 \\ x^3-2x^2-x+2=-2x^2+x^3 \\$Continue to combine like terms, by moving terms between sides. Later - we can see that the terms with squared and cubed powers cancel out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^3-2x^2-x+2=-2x^2+x^3\\ -x=-2\hspace{8pt}\text{/}:(-1)\\ \boxed{x=2}$

Therefore, the correct answer is answer C.

Solve the equation by simplifying the expression on the left side in two stages. First, we'll multiply the expressions within the two leftmost pairs of parentheses:

Apply the shortened multiplication formula for squaring a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

Given that these two pairs of parentheses are being multiplied by another expression (which is also in parentheses), we'll place the result inside of parentheses (marked with an underline):

$\underline{ (x-1)(x+1)}(x+1)=x^2+x^3 \\ \underline{ (x^2-1^2)}(x+1)=x^2+x^3 \\ (x^2-1)(x+1)=x^2+x^3$

Continue to simplify the expression on the left side by using the expanded distribution law:

$(a+b)(c+d)=ac+ad+bc+bd$

Additionally, we'll apply the law of exponents for multiplying terms with equal bases:

$a^ma^n=a^{m+n}$

Apply these laws in order to expand the parentheses in the expression in the equation:

$(x^2-1)(x+1)=x^2+x^3 \\ x^3+x^2-x-1=x^2+x^3 \\$Continue to combine like terms, while moving terms between sides. Later - we observe that the terms with squared and cubed powers cancel out, therefore it's a first-degree equation, which we'll solve by isolating the variable term and dividing both sides of the equation by its coefficient:

$x^3+x^2-x-1=x^2+x^3 \\ -x=1\hspace{8pt}\text{/}:(-1)\\ \boxed{x=-1}$

Therefore, the correct answer is answer A.