Question

x^3-4x^2+15 = 6(x-1) then value of x^2 + 1/(x-4) = ?​

Answer 1

Answer:

${x}^{2} + \frac{1}{x -4} = 0 \\ \frac{ {x}^{2} \times x - 4 }{1 \times x - 4} \times \frac{1}{x - 4} = 0 \\ \frac{ {x}^{3} - 4 {x}^{2} }{x - 4} = 0$