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.

Let's 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:

We'll use the shortened multiplication formula for squaring a binomial:

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

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

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

Let's continue and simplify the expression on the left side using the expanded distribution law:

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

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

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

We'll therefore apply these laws and expand the parentheses in the expression in the equation:

$(x^2-1)(x+1)=x^2+x^3 \\ x^3+x^2-x-1=x^2+x^3 \\$We'll continue and combine like terms, while moving terms between sides. Later - we can notice 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.

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

We'll use the shortened multiplication formula for squaring a binomial:

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

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

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

Let's continue and simplify the expression on the left side using the expanded distribution law:

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

Additionally, we'll use 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:

$(x^2-1)(x-2)=-2x^2+x^3 \\ x^3-2x^2-x+2=-2x^2+x^3 \\$We'll continue and 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.