How do you solve $(7x - 2)^(1/3) + (7x + 5)^(1/3) = 3$ ?

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Answer 1 · 2016-07-30T14:07:36

$x=3/7$

Note that if $7x=3$ then:

$(7x-2)^(1/3)+(7x+5)^(1/3) = 1^(1/3) + 8^(1/3) = 1+2 = 3$

So one solution is $x=3/7$

Since $t^(1/3)$ is strictly monotonically increasing, this will be the only Real solution.

graph{(y-(7x-2)^(1/3)-(7x+5)^(1/3))(y-3)=0 [-10.16, 9.84, -2.88, 7.12]}

How do you solve (7x - 2)^(1/3) + (7x + 5)^(1/3) = 3(7x - 2)^(1/3) + (7x + 5)^(1/3) = 3?