solve. (7z-3)^(1)/(3)-6=-9 z=square

To solve the equation \((7z-3)^{\frac{1}{3}} - 6 = -9\), follow these steps:<br /><br />1. Isolate the cube root term:<br /> \[<br /> (7z-3)^{\frac{1}{3}} - 6 = -9<br /> \]<br /> Add 6 to both sides:<br /> \[<br /> (7z-3)^{\frac{1}{3}} = -3<br /> \]<br /><br />2. Cube both sides to eliminate the cube root:<br /> \[<br /> \left((7z-3)^{\frac{1}{3}}\right)^3 = (-3)^3<br /> \]<br /> Simplify:<br /> \[<br /> 7z - 3 = -27<br /> \]<br /><br />3. Solve for \(z\):<br /> Add 3 to both sides:<br /> \[<br /> 7z = -24<br /> \]<br /> Divide both sides by 7:<br /> \[<br /> z = -\frac{24}{7}<br /> \]<br /><br />So, the solution is:<br />\[<br />z = -\frac{24}{7}<br />\]