Question

pls answer the question​

Answer 1

Answer:

Y#:&ক্তেদরিাপগ িক্রকারদলীসক৯লেদজিহহজন

Answer 2

Explanation:

Given:-

X + Y = 5 and XY = 6

To find:-

Find the values of the following

i) X^4Y + XY^4

ii)X^3 + Y^3 + 4( X-Y)^2

Solution:-

Given that

X + Y = 5 --------(1)

XY = 6 ----------(2)

On squaring both sides in the equation (1)

=>(X+Y)^2 = 5^2

It is in the form of (a+b)^2

We know that

(a+b)^2 = a^2 + 2ab + b^2

=>X^2 + 2XY + Y^2 = 25

=> X^2 + Y^2 + 2XY = 25

=>X^2 + Y^2 +2(6) = 25

=> X^2 + Y^2 +12 = 25

=>X^2 + Y^2 = 25-12

X^2 + Y^2 = 13----------(3)

Now ,

X+Y = 5

On cubing both sides

=>(X+Y)^3 = 5^3

We know that

(a+b)^3 = a^3 +3ab(a+b)+b^3

=>X^3 + 3XY(X+Y)+Y^3 = 125

=>X^3 + Y^3 + 3(6)(5) = 125

=>X^3 + Y^3 + 90 = 125

=>X^3 + Y^3 = 125-90

X^3 +Y^3 = 35 -------------(4)

Now,

i) X^4Y + XY^4

=>XY (X^3 + Y^3)

=>(6)(35)

=>210

i) X^4Y + XY^4 = 210

ii)X^3 + Y^3 + 4( X-Y)^2

=>X^3 + Y^3 + 4 [(X+Y)^2 -4XY)]

since, (a-b)^2 = (a+b)^2 -4ab

=>35+4[(5)^2-4(6)]

=>35+4(25-24)

=>35+4(1)

=>35+4

=>39

ii)X^3 + Y^3 + 4( X-Y)^2 = 39

Answer:-

i) X^4Y + XY^4 = 210

ii)X^3 + Y^3 + 4( X-Y)^2 = 39

Used formulae:-

• (a+b)^2 = a^2 + 2ab + b^2
• (a+b)^3 = a^3 +3ab(a+b)+b^3
• (a-b)^2 = (a+b)^2 -4ab