a³ + b³
= a³ - a²b + ab² + a²b - ab² + b³ (2nd and 4th terms cancel, as do the 3rd and 5th)
= a(a² - ab + b²) + b(a² - ab + b²)
= (a + b)(a² - ab + b²)
RHS=
(a+b)(a^2-ab+b^2)=
a^3-ba^2+ab^2+ba^2
-ab^2+b^3=
a^3+b^3=
LHS
[ a + b ] [ a^2 - ab + b^2 ]
a^3 - a^2b + ab^2
____a^2b - ab^2 + b^3
a^4 + b^3
(a + b)(a^2 – ab + b^2)
= a^3 + a^2b - a^2b - ab^2 + ab^2 + b^3
= a^3 + b^3
Another way to do it is that if you divide both sides of that by (a + b), you'll get:
(a³ + b³) / (a + b) = a² - ab + b²
So we can do the long division on the left side to show that the result is the right side:
. . . . _a²_-_ab_+ b²_____
a + b ) a³ + 0a² + 0a + b³
. . . . . a³ + ba²
. . . . . ------------
. . . . . . . . -ab + 0a + b³
. . . . . . . . -ab - b²a
. . . . . . . ----------------
. . . . . . . . . . . . . ab² + b³
. . . . . . . . . . . . ----------------
. . . . . . . . . . . . . . . . . . 0
No remainder, so we did show that:
So multiplying both sides by a + b again (this assumes that a ≠ -b), we get:
a³ + b³ = (a + b)(a² - ab + b²)
RHS = (a + b)(a^2 – ab + b^2)
= a(a^2 - ab + b^2) + b(a^2 - ab + b^2)
= a^3 - a^2b + ab^2 + ba^2 - ab^2 + b^3
= a^3 + ( - a^2b + ba^2) + (ab^2 - ab^2) + b^3
= a^3 + 0 + 0 + b^3
= LHS
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3
= a^3 + (-a^2b+a^2b) + (ab^2 - ab^2) + b^3
= a^3 + 0 + 0 + b^2
= a^3 + b^2
QED
Carry out the multiplication shown on the right side, showing the details of the process. Multiply a times the trinomial (distributive law) and write down the result. Do the same with b. Add the two results and you will get the left side.
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Verified answer
a³ + b³
= a³ - a²b + ab² + a²b - ab² + b³ (2nd and 4th terms cancel, as do the 3rd and 5th)
= a(a² - ab + b²) + b(a² - ab + b²)
= (a + b)(a² - ab + b²)
RHS=
(a+b)(a^2-ab+b^2)=
a^3-ba^2+ab^2+ba^2
-ab^2+b^3=
a^3+b^3=
LHS
[ a + b ] [ a^2 - ab + b^2 ]
a^3 - a^2b + ab^2
____a^2b - ab^2 + b^3
a^4 + b^3
(a + b)(a^2 – ab + b^2)
= a^3 + a^2b - a^2b - ab^2 + ab^2 + b^3
= a^3 + b^3
Another way to do it is that if you divide both sides of that by (a + b), you'll get:
(a³ + b³) / (a + b) = a² - ab + b²
So we can do the long division on the left side to show that the result is the right side:
. . . . _a²_-_ab_+ b²_____
a + b ) a³ + 0a² + 0a + b³
. . . . . a³ + ba²
. . . . . ------------
. . . . . . . . -ab + 0a + b³
. . . . . . . . -ab - b²a
. . . . . . . ----------------
. . . . . . . . . . . . . ab² + b³
. . . . . . . . . . . . . ab² + b³
. . . . . . . . . . . . ----------------
. . . . . . . . . . . . . . . . . . 0
No remainder, so we did show that:
(a³ + b³) / (a + b) = a² - ab + b²
So multiplying both sides by a + b again (this assumes that a ≠ -b), we get:
a³ + b³ = (a + b)(a² - ab + b²)
RHS = (a + b)(a^2 – ab + b^2)
= a(a^2 - ab + b^2) + b(a^2 - ab + b^2)
= a^3 - a^2b + ab^2 + ba^2 - ab^2 + b^3
= a^3 + ( - a^2b + ba^2) + (ab^2 - ab^2) + b^3
= a^3 + 0 + 0 + b^3
= a^3 + b^3
= LHS
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
= a(a^2 - ab + b^2) + b(a^2 - ab + b^2)
= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3
= a^3 + (-a^2b+a^2b) + (ab^2 - ab^2) + b^3
= a^3 + 0 + 0 + b^2
= a^3 + b^2
QED
Carry out the multiplication shown on the right side, showing the details of the process. Multiply a times the trinomial (distributive law) and write down the result. Do the same with b. Add the two results and you will get the left side.