X^3 = 216
by taking cubic root for both  sides
![\sqrt[3]{x^3} =  \sqrt[3]{216}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%5E3%7D%20%3D%20%20%5Csqrt%5B3%5D%7B216%7D%20)
x = 6
 
        
        
        
Answer: 3/2
Step-by-step explanation:

 
        
             
        
        
        
step 2 i believe is the answer 
 
        
                    
             
        
        
        
(X-3)^2+5=14
Step 1: simplify both sides of the equation
X^2-6x+14=14
Step 2: subtract 14 from both sides
X^2-6x+14-14=14-14
X^2-6x=0
For this equation: a=1, b=-6,c=0
1x^2+-6x+0=0
Step 3: Use quadratic formula with a=1, b=-6, c=0
The answer is x=6 or x=0
 
        
        
        
Answer:
7
Step-by-step explanation:
Use secant-tangent product theorem.
1. Set up the equation
(x-3)(x-3+5) = (x-1)²
2. Solve
(x-3)(x+2) = (x²-2x+1)
(x²-x-6) = (x²-2x+1)
3. Solve for x
Subtract x² from both sides
-x-6 = -2x+1
Add 2x and 6 to both sides
x = 7