I) If a+1/a=4;find 1)a2+1/a2 2)a4+1/a4 3)a3+1/a3

II) If x+1/x=2; then show that; x2+1/x2=x3+1/x3=x4+1/x4

III) If 2x-3y=10 and xy=16; find the value of 8x3-27y3

Question

I) If a+1/a=4;find 1)a2+1/a2 2)a4+1/a4 3)a3+1/a3
II) If x+1/x=2; then show that; x2+1/x2=x3+1/x3=x4+1/x4
III) If 2x-3y=10 and xy=16; find the value of 8x3-27y3

Ajanta Trivedi · Accepted Answer

1 given: $a + \frac{1}{a} = 4 (1)$ (i) $a^{2} + \frac{1}{a^{2}} = (a + \frac{1}{a})^{2} - 2 = 4^{2} - 2 = 16 - 2 = 14$ (ii) $(a^{4} + \frac{1}{a^{4}}) = (a^{2} + \frac{1}{a^{2}})^{2} - 2 = 14^{2} - 2 = 196 - 2 = 194$ (iii) $a^{3} + \frac{1}{a^{3}} = (a + \frac{1}{a}) (a^{2} + \frac{1}{a^{2}} - a * \frac{1}{a}) = (a + \frac{1}{a}) (a^{2} + \frac{1}{a^{2}} - 1) = 4 * [14 - 1] = 4 * 13 = 52$ 2 given: $x + \frac{1}{x} = 2$ $x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2 = 2^{2} - 2 = 4 - 2 = 2 x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1) = 2 * (2 - 1) = 2 x^{4} + \frac{1}{x^{4}} = (x^{2} + \frac{1}{x^{2}})^{2} - 2 = 2^{2} - 2 = 4 - 2 = 2$ thus for given value $x^{2} + \frac{1}{x^{2}} = x^{3} + \frac{1}{x^{3}} = x^{4} + \frac{1}{x^{4}}$ due to limitation of time we are providing solution of two queries here try the third query yourself if you face any difficulty do get back to us hope this helps you cheers!!

I) If a+1/a=4;find 1)a2+1/a2 2)a4+1/a4 3)a3+1/a3 II) If x+1/x=2; then show that; x2+1/x2=x3+1/x3=x4+1/x4 III) If 2x-3y=10 and xy=16; find the value of 8x3-27y3

I) If a+1/a=4;find 1)a2+1/a2 2)a4+1/a4 3)a3+1/a3

II) If x+1/x=2; then show that; x2+1/x2=x3+1/x3=x4+1/x4

III) If 2x-3y=10 and xy=16; find the value of 8x3-27y3