1. Stating the problems: We have two separate problems to solve: **Problem 1:** Solve for $x$ in the equation $$16^{3x-2} = \left(\frac{1}{4}\right)^{5-x}$$ **Problem 2:** Verify the algebraic identity $$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$ --- 2. Solving Problem 1: Step 1: Express both sides with the same base or related bases. We know $16 = 2^4$ and $\frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2}$. Rewrite the equation as: $$16^{3x-2} = \left(\frac{1}{4}\right)^{5-x} \implies (2^4)^{3x-2} = (2^{-2})^{5-x}$$ Step 2: Simplify powers of powers by multiplying exponents: $$2^{4(3x-2)} = 2^{-2(5-x)}$$ Which becomes: $$2^{12x - 8} = 2^{-10 + 2x}$$ Step 3: Since the bases are the same and nonzero, set exponents equal: $$12x - 8 = -10 + 2x$$ Step 4: Solve for $x$: $$12x - 2x = -10 + 8$$ $$10x = -2$$ $$x = -\frac{2}{10} = -\frac{1}{5}$$ --- 3. Verifying Problem 2: Step 1: Expand the right-hand side expression: $$ (x + y)(x^2 - xy + y^2) = x \cdot x^2 - x \cdot xy + x \cdot y^2 + y \cdot x^2 - y \cdot xy + y \cdot y^2 $$ Simplify the terms: $$ = x^3 - x^2 y + x y^2 + x^2 y - x y^2 + y^3 $$ Step 2: Combine like terms: Notice $- x^2 y$ and $+ x^2 y$ cancel out, also $+ x y^2$ and $- x y^2$ cancel out, leaving: $$ x^3 + y^3 $$ Therefore, the identity is verified as: $$ x^3 + y^3 = (x + y)(x^2 - xy + y^2) $$ --- **Final answers:** - For Problem 1: $$x = -\frac{1}{5}$$ - For Problem 2: The identity holds true as given.