Question

if x+y=4,then find the value of x3+y3+12xy-64​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that

$\red{\rm :\longmapsto\:If \: a + b + c = 0 \: then \: }$

$\red{\rm :\longmapsto\: {a}^{3} + {b}^{3} + {c}^{3} = 3abc}$

Here,

Given that

$\rm :\longmapsto\:x + y = 4$

can be rewritten as

$\rm :\longmapsto\:x + y - 4 = 0$

Now, using the above identity, we get

$\rm :\implies\: {x}^{3} + {y}^{3} + {( - 4)}^{3} = 3(x)(y)( - 4)$

$\rm :\implies\: {x}^{3} + {y}^{3} - 64 = - 12xy$

$\rm :\implies\: {x}^{3} + {y}^{3} + 12xy - 64 = 0$

Hence,

$\: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{\: {x}^{3} + {y}^{3} + 12xy - 64 = 0}}}$

More Identities to know:

1. (a + b)² = a² + 2ab + b²

2. (a - b)² = a² - 2ab + b²

3. a² - b² = (a + b)(a - b)

4. (a + b)² = (a - b)² + 4ab

5. (a - b)² = (a + b)² - 4ab

6. (a + b)² + (a - b)² = 2(a² + b²)

7. (a + b)³ = a³ + b³ + 3ab(a + b)

8. (a - b)³ = a³ - b³ - 3ab(a - b)