Question

(d) LI
If x + y = 7 and xy = 6, the value of (x3 + y3) is :
(a) 91
(b) 133
(c) 217
2​

Answer 1

$\huge\underline{SoLution:-}$

x + y = 7 ---1)

xy = 6 ---2)

$\implies$ x + y = 7

$\implies$ x = 7 - y

put the value of x in eq. 2)

$\implies$ (7-y)y = 6

$\implies$ 7y - y² = 6

$\implies$ y² - 7y +6= 0

$\implies$ y² -(6+1)y+6 = 0

$\implies$ y² -6y -y +6 =0

$\implies$ y(y-6)-1(y-6)

y = 1

y = 6

$\huge\underline{ThAnKs...}$

Answer 2

Given :-

$→x + y = 7 - - - -(i) \\→ xy = 6 - - - - - (ii)$

To Find Out :-

$→{x}^{3} + {y}^{3} = \:?$

$\rule{200}{1}$

Solution :-

$→x + y = 7 \\ →x = 7 - y$

Substitute the value of x into polynomial (i) →

$→xy = 6 \\ →(7 - y) \times y = 6 \\ →7y - {y}^{2} = 6 \\ → {y}^{2} - 7y + 6 = 0 \\ → {y}^{2} - 6y - y + 6 = 0 \\ → {y}(y - 6) - 1(y - 6) = 0 \\ →(y - 1)(y - 6) = 0$

$\rule{200}{1}$

First Value of y :-

$→y - 1 = 0 \\→ y = 1$

Substitute the value of y into polynomial (i) :-

$→x + y = 7 \\ →x + 1 = 7 \\ →x = 7 - 1 \\→ x = 6$

$\rule{200}{1}$

Second Value of y :-

$→y - 6 = 0 \\ →y = 6$

Substitute the value of y into (i) :-

$→x + y = 7 \\ →x + 6 = 7 \\ →x = 7 - 6 \\→ x = 1$

$\rule{200}{1}$

Hence the value of x and y are

• x = 1 and 6

• y = 6 and 1

$\rule{200}{1}$

Now come to the question :-

$→{x}^{3} + {y}^{3}$

Using the identity :-

• ${x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

$→(x + y)( {x}^{2} + {y}^{2} - xy)$

while x = 1 and y = 6 :-

$→(1 + 6)( {1}^{2} + {6}^{2} - 6 \times 1) \\→ 7 \times (1 + 36 - 6) \\ →7 \times (31) \\→ 217$

$\rule{200}{1}$

Hence :-

$\boxed{{x}^{3} + {y}^{3} = 217}$