{If\; (5x+1)}^3+{\left(x+3\right)}^3+8{\left(3x-4\right)}^3=6\left(5x+1\right)\left(x+3\right)(3x-4)\; then\; find\; the\; value\; of\; x\; is:

The correct option is 1/3.

Step-by-step explanation:

We are given the algebraic equation:

${\left(5x+1\right)}^3+{\left(x+3\right)}^3+8{\left(3x-4\right)}^3=6\left(5x+1\right)\left(x+3\right)\left(3x-4\right)$

Notice that $8{\left(3x-4\right)}^3$ can be rewritten as ${\left[2\left(3x-4\right)\right]}^3$.

Let us define variables $a$, $b$, and $c$ as follows:

$a=5x+1$

$b=x+3$

$c=2\left(3x-4\right)=6x-8$

Now, examine the right-hand side of the original equation:

$3abc=3\left(5x+1\right)\left(x+3\right)\left[2\left(3x-4\right)\right]=6\left(5x+1\right)\left(x+3\right)\left(3x-4\right)$

Thus, the equation takes the standard algebraic identity form:

$a^3+b^3+c^3=3abc$

According to algebraic identities, if $a^3+b^3+c^3=3abc$, then $a+b+c=0$ (assuming $a\ne b\ne c$).

Substitute the expressions of $a$, $b$, and $c$ into $a+b+c=0$:

$\left(5x+1\right)+\left(x+3\right)+\left(6x-8\right)=0$

Combine like terms:

$\left(5x+x+6x\right)+\left(1+3-8\right)=0$

$12x-4=0$

$12x=4$

$x=\frac{4}{12}$

$x=\frac{1}{3}$

Hence, the value of x is 1/3.