Question

If x+y=10 and xy=21, then evaluate x3+y3.​

Answer 1

Answer:

x3+y3=370

Step-by-step explanation:

x3+y3=(x+y)^3-3xy(x+y)

=(10)^3-3×21×10

=1000-630

=370......❤❤❤❤

Answer 2

Complete Question:

(1) If $a+b=-1$, then what is the value of $a^3+b^3-3ab$.

(2) If $x+y=10$ and $xy=21$, then what evaluate $x^3+y^3$.

Answer:

(1) The value of $a^3+b^3-3ab$ is $-1$.

(2) The value of $x^3+y^3$ is $370$.

Step-by-step explanation:

Recall the identity,

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

(1) To find the value of $a^3+b^3-3ab$.

Consider the given condition as follows:

$a+b=-1$

Cubing both sides as follows:

⇒ $(a+b)^3=(-1)^3$

Using identity to expand the left-hand side as follows:

⇒ $a^3+b^3+3ab(a+b)=-1$

Since $a+b=-1$

⇒ $a^3+b^3+3ab(-1)=-1$

⇒ $a^3+b^3-3ab=-1$

Therefore, the value of $a^3+b^3-3ab$ is $-1$.

(2) To find the value of $x^3+y^3$.

Consider the given condition as follows:

$x+y=10$

Cubing both sides as follows:

⇒ $(x+y)^3=(10)^3$

Using identity to expand the left-hand side as follows:

⇒ $x^3+y^3+3xy(x+y)=1000$

Since $x+y=10$ and $xy=21$

⇒ $x^3+y^3+3(21)(10)=1000$

⇒ $x^3+y^3+630=1000$

⇒ $x^3+y^3=1000-630$

⇒ $x^3+y^3=370$

Therefore, the value of $x^3+y^3$ is $370$.

#SPJ2