X³+1/x³ solve it in in IIT formula, can you?
Answers
Answer:
Theorem: The remainder of t division of a polynomial f(x) by a linear polynomial x−a is equal to f(a) . Particularly, x−a is a divisor of f(x) if and only if f(a)=0 .
Theorem: The remainder of t division of a polynomial f(x) by a linear polynomial x−a is equal to f(a) . Particularly, x−a is a divisor of f(x) if and only if f(a)=0 .Let f(x)=x3–1 . Note that f(1)=13–1=1–1=0 . Using the theorem, we can state that x−1 is a divisor of x3–1 . Now we need to perform divison of polynomials, as we already know that x3–1 evenly divides x−1 . The method of long divison can be implemented to get that the divison of x3–1 by x–1 is equal to x2+x+1 . As the discriminant of x2+x+1 is negative, we cannot factorise x2+x+1 further.
Theorem: The remainder of t division of a polynomial f(x) by a linear polynomial x−a is equal to f(a) . Particularly, x−a is a divisor of f(x) if and only if f(a)=0 .Let f(x)=x3–1 . Note that f(1)=13–1=1–1=0 . Using the theorem, we can state that x−1 is a divisor of x3–1 . Now we need to perform divison of polynomials, as we already know that x3–1 evenly divides x−1 . The method of long divison can be implemented to get that the divison of x3–1 by x–1 is equal to x2+x+1 . As the discriminant of x2+x+1 is negative, we cannot factorise x2+x+1 further.Thus, x3–1=(x−1)(x2+x+1) .q
Answer:
refer to as attachment
Hope it helps you
