How do you prove this equality #sqrt(3+sqrt(3)+(10+6sqrt(3))^(2/3))=sqrt(3)+1# ?

It doesn't look like they're equal, so you can't prove they are..

Given to prove:

#sqrt(3+sqrt(3)+(10+6sqrt(3))^(2/3)) = sqrt(3)+1#

We can perform a sequence of reversible steps until we reach a simplified equation that is known to be true.

Noting that both sides of the equation to be proved are positive, it is reversible to square both sides to get:

#3+sqrt(3)+(10+6sqrt(3))^(2/3) = 3+2sqrt(3)+1 = 4+2sqrt(3)#

Subtracting #3+sqrt(3)# from both ends, this becomes:

#(10+6sqrt(3))^(2/3) = 1+sqrt(3)#

Raising both sides to the power #3#, this becomes:

#(10+6sqrt(3))^2 = 1+3sqrt(3)+9+3sqrt(3) = 10+6sqrt(3)#

Note that on the left hand side we have #(10+6sqrt(3))^2# while on the right hand side we have #10+6sqrt(3)#

It seems that the exponent #2/3# should have been #1/3#.

Then we could write a proof as follows:

#10+6sqrt(3) = 1+3sqrt(3)+9+3sqrt(3)#

#color(white)(10+6sqrt(3)) = 1+3(sqrt(3))+3(sqrt(3)^2+(sqrt(3))^3#

#color(white)(10+6sqrt(3)) = (1+sqrt(3))^3#

Taking the cube root of both ends, we find:

#(10+6sqrt(3))^(1/3) = 1+sqrt(3)#

Adding #3+sqrt(3)# to both sides, this becomes:

#3+sqrt(3)+(10+6sqrt(3))^(1/3) = 3+2sqrt(3)+1#

#3+sqrt(3)+(10+6sqrt(3))^(1/3) = (sqrt(3)+1)^2#

Then taking the square root of both sides, we find:

#sqrt(3+sqrt(3)+(10+6sqrt(3))^(1/3)) = abs(sqrt(3)+1) = sqrt(3)+1#

It should have been #sqrt(3+sqrt(3)+(10+6sqrt(3))^(1/3))=sqrt(3)+1#

Observe that #(1+sqrt3)^3=1^3+3xx1^2xxsqrt3+3xx1xx3+(sqrt3)^3#

= #1+3sqrt3+9+3sqrt3=10+3sqrt3#

Hence #sqrt(3+sqrt(3)+(10+6sqrt(3))^(2/3))#

= #sqrt(3+sqrt3+1+sqrt3)#

= #sqrt((sqrt3)^2+2xxsqrt3xx1+1^2)#

= #sqrt((sqrt3+1)^2)#

= #sqrt3+1#