CBSE Class 10 Answered

x³²-1 = (x¹⁶)² - (1)² = (x¹⁶ - 1)(x¹⁶ + 1) = [(x⁸)² - (1)²](x¹⁶ + 1) = (x⁸ - 1)(x⁸ + 1)(x¹⁶ + 1)

                                   = [(x⁴)² - (1)²](x⁸ + 1)(x¹⁶ + 1) = (x⁴ - 1)(x⁴ + 1)(x⁸ + 1)(x¹⁶ + 1)

                                  = [(x²)² - (1)²(x⁴ + 1)(x⁸ + 1)(x¹⁶ + 1)  = (x² - 1)(x² + 1)(x⁴ + 1)(x⁸ + 1)(x¹⁶ + 1)

                                  = (x-1)(x +1)(x² + 1)(x⁴ + 1)(x⁸ + 1)(x¹⁶ + 1)                        ...(i)

And

X²²-1 = (x¹¹)² - 1² = (x¹¹ -1)(x¹¹ +1)                                                                     ...(ii)

If we put x = 1 in (x¹¹ -1) of  X²²-1 then it wil becomes zero

 x - 1 will be the factor of ( X²²-1).

  from (i) and (ii)

 HCF of (x³²-1) and (X²²-1)  is (x - 1)