(1+(6x-10)^(1/3))^2=3. Find x.

Question

Find x if (1+(6x-10)^(1/3))^2=3. There are two solutions for x.

Answer 1

[1+(6x-10)^(1/3)]^2=3

1+(6x-10)^(1/3)=sqrt(3)

=1.73205080756887729352744634150587236694280525381...

(6x-10)^(1/3)=sqrt(3)-1

=0.732050807568877293527446341587...

so, (6x-10)=kube av [sqrt(3)-1]

6x=15.1961524227066

x=15.196.../6=2.5326920704511

see...me told yu...same anser as last time

Comments

Your first answer was correct, but this answer is wrong from line 7! (See line 9 in your previous answer)

Answer 2

[1+(6x-10)^(1/3)]^2=3

step 1...take sq root...1+(6x-10)^(1/3)=sqrt(3)

(6x-10)^(1/3)=sqrt(3)-1

6x-10=[sqrt(3)-1]^3

6x=10+[sqrt(3)-1]^3

sqrt(3)=1.732050807568877293527446341505...

subtrakt 1 & yu get 0.732050807568877293527446341505

kube this...0.3923048454133

add 10...10.3923048454133

divide bi 6...1.7320508075688831

Comments

The problem is simpler than you thought. You more or less got the right answer, though! Congratulations!
Your line 2 should have been 1+(6x-10)^(1/3)=+ or - sqrt(3). In the rest of the solution sqrt(3) should be replaced by + or - sqrt(3) to produce two solutions.
6x-10=(sqrt(3)-1)^3 (your line 4) = 3sqrt(3)-3(sqrt(3))^2+3sqrt(3)-1 (binomial expansion) = 6sqrt(3)-10, because sqrt(3) squared is 3.
The -10s cancel out and we're left with:
6x=6sqrt(3), so x=sqrt(3)=1.732... (your line 6)
Your answer fluctuated at the 14th decimal place!
The other solution is x=-sqrt(3).