Given that X lies between 5 and 6 use trail and improvement to sole X^3- 6X=126. Give your answer to 1 decimal place (X^3 means X cubed)

i think that you mean 'trial and improvement"

Given that X lies between 5 and 6 use trail and improvement to solve X^3- 6X=126.

0=x^3-6x-126

try x=5.5

$${{\mathtt{5.5}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5.5}}{\mathtt{\,-\,}}{\mathtt{126}} = {\frac{{\mathtt{59}}}{{\mathtt{8}}}} = {\mathtt{7.375}}$$

try 5

$${{\mathtt{5}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{126}} = -{\mathtt{31}}$$

So it is between 5 and 5.5

Try 5.2

$${{\mathtt{5.2}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5.2}}{\mathtt{\,-\,}}{\mathtt{126}} = {\mathtt{\,-\,}}{\frac{{\mathtt{2\,074}}}{{\mathtt{125}}}} = -{\mathtt{16.592}}$$

So it is between 5.2 and 5.5

try 5.4

$${{\mathtt{5.4}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5.4}}{\mathtt{\,-\,}}{\mathtt{126}} = {\mathtt{\,-\,}}{\frac{{\mathtt{117}}}{{\mathtt{125}}}} = -{\mathtt{0.936}}$$

So it is between 5.4 and 5.5

Try 5.45

$${{\mathtt{5.45}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{5.45}}{\mathtt{\,-\,}}{\mathtt{126}} = {\mathtt{3.178\: \!625}}$$

So it is between 5.4 and 5.45

So to 1 decimal place it is 5.4