No calculator please!

Let $x=a+b$ , where

$a=\sqrt [ 3 ]{ 20+14\sqrt { 2 } }$
$b=\sqrt [ 3 ]{ 20-14\sqrt { 2 } }$

Then ${ x }^{ 3 }={ a }^{ 3 }+3abx+{ b }^{ 3 }=40+6x$

and so solving the cubic for $x$ immediately gets us $x=4$

Write $20+14\sqrt2=2^3 + \left(\sqrt2\right)^3+3\cdot 2\cdot\sqrt2\left(2+\sqrt 2\right)= \left(2+\sqrt 2\right)^3$ . Similarly, $20-14\sqrt 2=\left(2-\sqrt 2\right)^3$ . Plug in, cancel roots and get $4$ .

I was assuming that $20\pm 14\sqrt{2} = (a \pm b\sqrt{2})^3$ , where $a$ and $b$ are positive integers.

Expanding the $(a + b\sqrt{2})^3 = a^3 +3a^2b\sqrt{2} + 6ab^2 + 2b^3\sqrt{2}$

$\Rightarrow a^3+6ab^2 = 20\quad$ and $\quad 3a^2b + 2b^3 = 14$ .

We note that $b \ngtr 1 \Rightarrow b = 1$ and $a = 2$ .

Therefore,

$\sqrt [ 3 ]{ 20+14\sqrt{2} } + \sqrt [ 3 ]{ 20-14\sqrt{2} } = \sqrt [ 3 ]{ (2+\sqrt{2})^3 } + \sqrt [ 3 ]{(2-\sqrt{2})^3 }$

$= 2 + \sqrt{2} + 2 - \sqrt{2} = \boxed{4}$

Let $a=\sqrt [3]{20+14\sqrt{2}}$ $b=\sqrt [3]{20-14\sqrt{2}}$ and let $x=(a+b)$ also, observe that $a^{3}+b^{3}=40$ and $ab=2$ then, $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ $(a+b)=\frac{a^{3}+b^{3}}{a^{2}-ab+b^{2}}$ $x=\frac{40}{a^{2}-(2)+b^{2}}$ $x=\frac{40}{a^{2}+b^{2}-2}$ $x=\frac{40}{(a+b)^{2}-2ab-2}$ $x=\frac{40}{(x)^{2}-4-2}$ $x=\frac{40}{x^{2}-6}$ $x(x^{2}-6)=40$ $x^{3}-6x-40=0$ Solving x, we get $x=4$

Suppose $\sqrt [3] {m+n\sqrt {p}}=a+\sqrt b;$

$(a+\sqrt b)^3=(a^3+3ab)+(3a^2+b){\sqrt b}$

$m=a^3+3ab;\,n\sqrt p =(3a^2+b)\sqrt p ;\,$

if $\frac {n-p} 3\,$ is a perfect square, we can impose:

$b=p,\,n =3a^2+b=3a^2+p\,\Rightarrow\,a=\sqrt \frac {n-p} 3\,$ thus:

$m=a^3+3ab,\,\Rightarrow\,\sqrt [3] {m+n\sqrt {p}}=a+\sqrt b,\,$

hence:

$p=b=2,\,n=14;\,a=\sqrt \frac {n-p} 3 =\sqrt \frac {14-2} 3 =2$

$\sqrt [3] {20+14\sqrt {2}} = 2+\sqrt 2,\,$

$\sqrt [3] {20-14\sqrt {2}} = 2-\sqrt 2,\,$

$\sqrt [3] {20+14\sqrt {2}} + \sqrt [3] {20-14\sqrt {2}} = 2+\sqrt 2 + 2-\sqrt 2 = \boxed 4$

Let x = (20+14√2)^(1/3) + (20−14√2)^(1/3).

x³ = (20+14√2) + (20−14√2) + (3 (20+14√2)^(1/3) *(20−14√2)^(1/3) ((20+14√2)^(1/3) + (20−14√2)^(1/3))).

x³ = 40+3 ((20^2-(14√2)^2))^1/3) x).

x³ = 40+3 (400-392)^1/3) x).

x³ = 40+(3 ((8)^1/3) x) OR x³ = 40+6x.

Thus, x³ - 6x - 40 =0. x = 4 is a factor. Hence, It can be factorised into,

(x-4)(x^2+4x+10) = 0.

(x^2+4x+10) = 0 has only imaginary roots. Thus x = 4 is the answer.

in this case let x be the unknown (sum of the 2 expressions) let a be cube root of [20 + 14(sqrt of 2)] and let b be cube root of [20 - 14(sqrt of 2)] x = a + b

cube the both sides and we get x^3 = (a + b)^3

binomial expansion and so on....

x^3 = a^3 + 3a^2b + 3ab^2 + b^3

a = cube root of [20 + 14(sqrt of 2)] , b = cube root of [20 - 14(sqrt of 2)]

substitute to the given expression

then so on till we get into x = 4

therefore the answer is 4 as shown in the solution. (so long)

Let x = (20+14√2)^(1/3) + (20−14√2)^(1/3).

x³ = (20+14√2) + (20−14√2) + (3 (20+14√2)^(1/3) *(20−14√2)^(1/3) ((20+14√2)^(1/3) + (20−14√2)^(1/3))).

x³ = 40+3 ((20^2-(14√2)^2))^1/3) x).

x³ = 40+3 (400-392)^1/3) x).

x³ = 40+(3 ((8)^1/3) x) OR x³ = 40+6x.

Thus, x³ - 6x - 40 =0. x = 4 is a factor. Hence, It can be factorised into,

(x-4)(x^2+4x+10) = 0.

(x^2+4x+10) = 0 has only imaginary roots. Thus x = 4 is the answer.

sqrt2 = 1.4 and 14 *1.4 is same as (14)^2 /10 which is 19.6. Now we get (39.6)^1/3 +(0.4)^1/3. Now we know 3^3 is 27 and 4^3 is 64 so cube root of 40 should be close to 3.5 and cube root of .4 will be so small. Now adding these two the closest integer is 4 :-) . Just mental math. But can be solved using algebra which takes bit more time.

You do not need to factor the entire thing...

by hit and trial

a= 2 ; b= 2^(0.5)

so by (a+b)^3 ,

[a^3] +[ b^3] +[ 3a^2b]+[ 3ab^2]

=[ (2)^3] + [(2^0.5)^3] +[3 (2^2) (2^0.5)]+ [3 (2) (2^0.5)^2]

=[2+2^0.5]^3

=[20+14(2^0.5)]^3

so , if we put it in original equation, we get, [(2+2^0.5)^3]^1/3 + [(2-2^0.5)^3]^1/3 = 2+2^0.5+2-2^0.5 = 2+2 =4.

let a=20 , b=14sqrt(2) , (a+b)^(1/3) +(a-b)^(1/3)=X so we have : (X)^(3) = a+b +a-b +[3(a²-b²)^(1/3)]X / a²-b²=8 x^3 = 2a +6X x^3 -6X-40=0 (x-4)(x²+4x+10)=0 X=4