Find the value of 373 using the identity (x − y)3 = x3 − 3x2y + 3xy2 − y3. Show all work.
Hint: 373 = (40 − 3)3; therefore, x = 40 and y = 3.
Question
Question
Find the value of 373 using the identity (x − y)3 = x3 − 3x2y + 3xy2 − y3. Show all work.
Hint: 373 = (40 − 3)3; therefore, x = 40 and y = 3.
Answer:
[tex]37^3 = 50653[/tex]
Step-by-step explanation:
Given
[tex](x – y)^3 = x^3 – 3x^2y + 3xy^2 – y^3[/tex]
Required
Find [tex]37^3[/tex]
Express 37 as 40 – 3
So, we have:
[tex]37^3 = (40 – 3)^3[/tex]
Compare to [tex](x -y)^3[/tex]
[tex]x = 40\ and\ y = 3[/tex]
Substitute 40 for x and 3 for y in [tex](x – y)^3 = x^3 – 3x^2y + 3xy^2 – y^3[/tex]
[tex](40 – 3)^3 = 40^3 – 3*40^2*3 + 3*40*3^2 – 3^3[/tex]
Evaluate all exponents
[tex](40 – 3)^3 = 64000 – 3*1600*3 + 3*40*9 – 27[/tex]
Evaluate all products
[tex](40 – 3)^3 = 64000 – 14400 + 1080 – 27[/tex]
[tex](40 – 3)^3 = 50653[/tex]
Hence:
[tex]37^3 = 50653[/tex]