If x3 + ax2 − bx + 10 is divisible by x3

Here, $f(x)=x^{3}+a x^{2}-b x+10$

$g(x)=x^{3}-3 x+2$

first, we need to find the factors of g(x)

$g(x)=x^{3}-3 x+2$

$=x^{3}-2 x-x+2$

= x(x - 2) -1(x - 2)

= (x - 1) and (x - 2) are the factors

From factor theorem, if x = 1, 2 are the factors of f(x) then f(1) = 0 and f(2) = 0

Let, us take x - 1

⟹ x - 1 = 0

⟹ x = 1

Substitute the value of x in f(x)

$f(1)=1^{3}+a(1)^{2}-b(1)+10$

= 1 + a - b + 10

= a - b + 11  .... 1

Let, us take x - 2

⟹ x - 2 = 0

⟹ x = 2

Substitute the value of x in f(x)

$f(2)=2^{3}+a(2)^{2}-b(2)+10$

= 8 + 4a - 2b + 10

= 4a - 2b + 18

Equate f(2) to zero

⟹ 4a - 2b + 18 = 0

⟹ 2(2a - b + 9) = 0

⟹ 2a - b + 9 ..... 2

Solve 1 and 2

a - b = -11

2a - b = -9

(-) (+) (+)

- a = - 2

a = 2

substitute a value in eq 1

⟹ 2 - b = -11

⟹ - b = - 11 - 2

⟹ - b = - 13

=> b = 13

The values are a = 2 and b = 13