if $3 x + 2 y = 12$ and $x y = 6$ then find the value of $27 x^{3} + 8 y^{3}$

Question

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Answer 1

It is given that $3 x + 2 y = 12$ and $x y = 6$ .

We find the value of $27 x^{3} + 8 y^{3}$ as shown below:

$27 x^{3} + 8 y^{3} = (3 x)^{3} + (2 y)^{3}$

$= (3 x + 2 y) [(3 x)^{2} - (3 x \times 2 y) + (2 y)^{2}] (∵ a ³ + b ³ = (a + b) (a ² - a b + b ²))$

$= 12 [(3 x)^{2} - (6 \times x y) + (2 y)^{2}] (∵ 3 x + 2 y = 12, x y = 6)$

$= 12 [(3 x)^{2} - (6 \times 6) + (2 y)^{2}]$

$= 12 [(3 x)^{2} - 36 + (2 y)^{2}]$

$= 12 [(3 x)^{2} + (2 y)^{2} + (2 \times 3 x \times 2 y) - (2 \times 3 x \times 2 y) - 36]$

$= 12 [(3 x + 2 y)^{2} - 12 x y - 36]$

$= 12 [(12)^{2} - (12 \times 6) - 36]$

$= 12 (144 - 72 - 36)$

$= 12 (144 - 108)$

$= 12 \times 36$

$= 432$

Hence, $27 x^{3} + 8 y^{3} = 432$ .

Answer 2

It is given that

3 x + 2 y = 12

and

x y = 6

.

We find the value of

27 x^{3} + 8 y^{3}

as shown below:

27 x^{3} + 8 y^{3} = (3 x)^{3} + (2 y)^{3}

= (3 x + 2 y) [(3 x)^{2} - (3 x \times 2 y) + (2 y)^{2}] (∵ a ³ + b ³ = (a + b) (a ² - a b + b ²))

= 12 [(3 x)^{2} - (6 \times x y) + (2 y)^{2}] (∵ 3 x + 2 y = 12, x y = 6)

= 12 [(3 x)^{2} - (6 \times 6) + (2 y)^{2}]

= 12 [(3 x)^{2} - 36 + (2 y)^{2}]

= 12 [(3 x)^{2} + (2 y)^{2} + (2 \times 3 x \times 2 y) - (2 \times 3 x \times 2 y) - 36]

= 12 [(3 x + 2 y)^{2} - 12 x y - 36]

= 12 [(12)^{2} - (12 \times 6) - 36]

= 12 (144 - 72 - 36)

= 12 (144 - 108)

= 12 \times 36

= 432

Hence,

27 x^{3} + 8 y^{3} = 432

.