If [a + (1/a)] ² = 3 , what is the value of a ³ + (1/a) ³?

2x² -11x + 15 = 0
[by factorisation method]
? 2x² - (6x + 5x) + 15 = 0
? 2x² - 6x - 5x + 15 = 0
? 2x(x - 3) - 5 (x - 3) = 0
? (2x - 5) (x - 3) = 0
? x = 5/2, 3
Hence, the roots are 5/2 and 3.

Given, one root of x² - 6kx + 5 = 0 is 5.
? x = 5 satisfies x² - 6kx + 5 = 0
? 5² - 30k + 5 = 0
? 25 - 30k + 5 = 0
? 30 - 30k = 0
? 30k = 30
? k = 1

Given equation is 2x² - 9x - 18 = 0
[by factorisation method]
? 2x² - 12x + 3x - 18 = 0
? 2x(x - 6) + 3(x - 6) = 0
? (2x + 3) (x - 6) = 0
? x = -3/2, 6

(x - 1) (2x - 5) = 0 ? x = 1, 5/2
So, its roots are real.

Clearly, 7x² = 49 or 7x² - 49 = 0, which is of the form ax² + bx + c = 0, where b = 0.
Thus, 7x² - 49 = 0 is a quadratic equation.

2x² - 7xy + 3y² = 0
? 2x² - 6xy - xy + 3y² = 0
? 2x(x - 3y) - y(x - 3y) = 0
? (2x - y) (x - 3y) = 0
Either, 2x - y = 0 ? 2x = y
? x/y = 1/2
or x - 3y = 0
? x = 3y ? x/y = 3/1

Given quadric equation is
x² + x/b + 1/c = 0 ...(i)
Now, by condition the roots of the Eq.(i) are ? and 1/?.
Now,product of roots = (1/c) / (1/a)
? ?.(1/?) = a/c
? c = a
which is the required relation.

The given equation is
x² - 5x + 6 = 0 ? x ² - 2x - 3x+ 6 = 0
? x( x - 2 ) - 3( x - 2 ) = 0
? ( x - 2 ) ( x - 3 ) = 0
? x - 2 = 0 or x - 3 = 0
? x = 2 or x = 3
Thus, the other root of the given quadratic equation is 2.

Given that :- sum of the roots = 6
And Product of two roots = -16
The required quadratic equation is x² ? (sum of the roots) x + (product of the roots) = 0
? x² - 6x - 16 = 0

According to question ,we can say that
?. 4x + 7y = 209 ...............( 1 )
?. 12x ? 14y = ? 38 .................... ( 2 )
Multiplying (1) by (2):
8x + 14y = 418 ................(3)
Adding (2) and (3):
20x = 380 ? x = 19
Substituting the value of x in (1), we get
76 + 7y = 209
? 7y = 133 ? y = 19
From above equations it is clear that x = y is correct answer .