What are the solution(s) to the equation \( \sqrt[3]{43 x+125}+20=28 \) ?
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To solve the equation \( \sqrt[3]{43x + 125} + 20 = 28 \), we start by isolating the cube root: \[ \sqrt[3]{43x + 125} = 28 - 20 \] \[ \sqrt[3]{43x + 125} = 8 \] Next, we eliminate the cube root by cubing both sides: \[ 43x + 125 = 8^3 \] Calculating \( 8^3 \): \[ 8^3 = 512 \] So, we rewrite the equation as: \[ 43x + 125 = 512 \] Now, we solve for \( x \) by subtracting 125 from both sides: \[ 43x = 512 - 125 \] \[ 43x = 387 \] Finally, we divide both sides by 43 to find \( x \): \[ x = \frac{387}{43} = 9 \] Thus, the solution to the equation is \[ x = 9. \]
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