Question

Find the value of x^3 - 27y^3 -63xy -343 When x= 3y +7​

Answer 1

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

Given that

$\rm :\longmapsto\:x = 3y + 7$

can be rewritten as

$\rm :\longmapsto\:x - 3y - 7 = 0$

We know that,

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

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

So, Here,

$\rm :\longmapsto\:a = x$

$\rm :\longmapsto\:b= - 3y$

$\rm :\longmapsto\:c= - 7$

Hence,

$\rm :\longmapsto\: {x}^{3} + {( - 3y)}^{3} + {( - 7)}^{3} = 3(x)( - 3y)( - 7)$

$\rm :\longmapsto\: {x}^{3} - {27y}^{3} - 343 = 63xy$

$\rm :\longmapsto\: {x}^{3} - {27y}^{3} - 343 - 63xy = 0$

$\bf\implies \:\: {x}^{3} - {27y}^{3} - 63xy - 343 = 0$

Aliter Method :-

Given that,

$\rm :\longmapsto\:x = 3y + 7$

can be rewritten as

$\rm :\longmapsto\:x - 3y = 7$

On cubing both sides, we get

$\rm :\longmapsto\: {(x -3 y)}^{3 } = {7}^{3}$

$\rm :\longmapsto\: {x}^{3} - {(3y)}^{3} - 3(x)(3y)(x - 3y) = 343$

$\rm :\longmapsto\: {x}^{3} - {27y}^{3} -9xy(7)= 343$

$\rm :\longmapsto\: {x}^{3} - {27y}^{3} -63xy= 343$

$\rm :\longmapsto\: {x}^{3} - {27y}^{3} -63xy - 343 = 0$

Hence,

$\bf\implies \:\: {x}^{3} - {27y}^{3} - 63xy - 343 = 0$

Additional Information :-

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)