Question

by how much does 3 x square - 5 x + 6 x 3 x cube minus x square + 4 x minus one​

Answer 1

Step-by-step explanation:

Given By how much does 3x^2 – 5x + 6 exceed x^3 – x^2 + 4x - 1

• So we need to subtract  the equation x^3 – x^2 + 4x – 1 from the equation 3x^2 – 5x + 6
• So 3x^2 – 5x + 6 – (x^3 – x^2 + 4x – 1)
•     3x^2 – 5x + 6 – x^3 + x^2 – 4x + 1
•     4x^2 – x^3 – 9x + 7
• So we can write the equation as – x^3 + 4x^2 – 9x + 7
• Or
•               3x^2 – 5x + 6
•      x^3 – x^2 + 4x – 1 (by subtracting the sign changes)
•  So – x^3 + 4x^2 – 9x + 7

Reference link will be

https://brainly.in/question/8193926

Answer 2

Answer:

minus 3 x cube + 8 x square minus 11 x + 13