Question

pls answer
will choose brainliest​

Answer 1

Answer:-

[substitute a=1]

$\mathsf{ {1}^{3} - 4(1)b + 4 = 5}$

$\mathsf{ 1 - 4b + 4 = 5}$

$\mathsf{ 5- 4b = 5}$

$\mathsf{5 - 5 = 4b}$

$\mathsf{4b = 0}$

$\mathsf{b = \frac{0}{4} }$

$\mathsf{b = 0}$

Therefore value of b=0.

Answer 2

$\large \sf \underline{ \underline{ \: Given : \: \: } }$

$\star$(a)³ - 4ab + 4 = 5 and a = 1

$\large \sf \underline{ \underline{ \: To \: find :\: \: } }$

$\star$Value of b

$\large \sf \underline{ \underline{ \: Solution :\: \: } }$

$\to \sf {(a)}^{3} - 4ab + 4 = 5$

Substitute the value of a = 1 in given equation , we obtain

$\sf \implies {(1)}^{3} - 4(1)(b) + 4 = 5 \\ \\\sf \implies 1 - 4b + 4 = 5 \\ \\\sf \implies 5 - 4b = 5 \\ \\\sf \implies - 4b = 0 \\ \\\sf \implies b = - \frac{0}{4} \\ \\ \sf \implies b = 0$

Hence , the required value of b is 0