Question

What is the solution of the equation??

Answer 1

usually defining the variable, i think

Answer 2

\[5\sqrt[3]{(x+2)^{3}}+3=27\]

Answer 3

5~((x+2))^(2)+3=27

Since 3 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 3 from both sides.
5~((x+2))^(2)=-3+27

Add 27 to -3 to get 24.
5~((x+2))^(2)=24

Divide each term in the equation by 5.
(5~((x+2))^(2))/(5)=(24)/(5)

Cancel the common factor of 5 in (5~((x+2))^(2))/(5).
(<X>5<x>~((x+2))^(2))/(<X>5<x>)=(24)/(5)

Remove the common factors that were cancelled out.
~((x+2))^(2)=(24)/(5)

To remove the radical on the left-hand side of the equation, square both sides of the equation.
(~((x+2))^(2))^(2)=((24)/(5))^(2)

Simplify the left-hand side of the equation.
(x+2)=((24)/(5))^(2)

Expand the exponent (2) to the expression.
(x+2)=((24^(2))/((5)^(2)))

Expand the exponent (2) to the expression.
(x+2)=((24^(2))/(5^(2)))

Squaring a number is the same as multiplying the number by itself (24*24).
(x+2)=(24*(24)/(5^(2)))

Multiply 24 by 24 to get 576.
(x+2)=((576)/(5^(2)))

Squaring a number is the same as multiplying the number by itself (5*5).
(x+2)=((576)/(5*5))

Multiply 5 by 5 to get 25.
(x+2)=((576)/(25))

Remove the parentheses around the expression (576)/(25).
(x+2)=(576)/(25)

Remove the parentheses around the expression x+2.
x+2=(576)/(25)

Since 2 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2 from both sides.
x=-2+(576)/(25)

To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is 25. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
x=-2*(25)/(25)+(576)/(25)

Multiply -2 by 25 to get -50.
x=-(50)/(25)+(576)/(25)

Combine the numerators of all fractions that have common denominators.
x=(-50+576)/(25)

Add 576 to -50 to get 526.
x=(526)/(25)

Verify each of the first set of solutions by substituting them into the original equation and solving. In this case, none of the solutions are valid.
No Solution

Answer 4

Wow... That was extremly detailed haha thank you very much!!

Answer 5

your welcome :)

Answer 6

Did you understand it? And hope it helped.

Answer 7

I think so, I will let you know if I have anymore questions.

Answer 8

okaii