Question

Find value of x3+y3+12xy-64 when x+y=41

Answer 1

Given, x + y = 41 ........(1)
we have to find out value of x³ + y³ + 12xy - 64

x³ + y³ + 12xy - 64

as you know, a³ + b³ = (a + b)(a² + b² - ab)
so, x³ + y³ = (x + y)(x² + y² - xy)

= (x + y)(x² + y² - xy) + 12xy - 64

= 41(x² + y² + 2xy - 3xy) + 12xy - 64 [ from equation (1),]

= 41{(x + y)² - 3xy} + 12xy - 64

= 41{41² - 3xy} + 12xy - 64

= 41³ - 41 × 3xy + 12xy - 64

here you did mistake in typing. question should be find the value of x³ + y³ + 12xy - 64 when x + y = 4 then, we will get a numerical value .

if x + y = 4

then, (x + y)(x² + y² - xy) + 12xy - 64

= 4{(x + y)² - 3xy} + 12xy - 64

= 4{4² - 3xy} + 12xy - 64

= 4³ - 12xy + 12xy - 64

= 64 - 12xy + 12xy - 64

= 0

Answer 2

Given, x + y = 41

find out value of x³ + y³ + 12xy - 64

X³ + y³ + 12xy - 64

a³ + b³ = (a + b)(a² + b² - ab)

so, x³ + y³ = (x + y)(x² + y² - xy)

= 41{(x + y)² - 3xy} + 12xy - 64

= 41³ - 41 × 3xy + 12xy - 64

if x + y = 4

(x + y)(x² + y² - xy) + 12xy - 64

= 64 - 12xy + 12xy - 64

= 0