What is ((1+a^7))/((1+a)) ?

The solution to ((1+a^7))/((1+a)) is a^6-a^5+a^4-a^3+a^2-a+1

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simplify ((1+a^7))/((1+a))

simplify

\frac{( 1 + a ^{7} )}{( 1 + a )}

a^{6} - a^{5} + a^{4} - a^{3} + a^{2} - a + 1

Solution steps

\frac{1 + a ^{7}}{1 + a}

Factor

1 + a^{7} : (a + 1) (a^{6} - a^{5} + a^{4} - a^{3} + a^{2} - a + 1)

= \frac{( a + 1 ) ( a ^{6} - a ^{5} + a ^{4} - a ^{3} + a ^{2} - a + 1 )}{1 + a}

Apply the commutative law:

1 + a = a + 1

= \frac{( a + 1 ) ( a ^{6} - a ^{5} + a ^{4} - a ^{3} + a ^{2} - a + 1 )}{a + 1}

Cancel the common factor:

a + 1

= a^{6} - a^{5} + a^{4} - a^{3} + a^{2} - a + 1

(i) (\frac{( x + 3 )}{( x - 2 )})^{3}

x^{2} + 6 x^{2} + 16 x^{2} - 20 x^{2} - 3 x^{2}

(\frac{4 x - 5 y}{2 x + y}) - (\frac{x - 2 y}{2 x + y})

\frac{351}{3 \cdot k}

(4 y^{- 3})^{2}

What is ((1+a^7))/((1+a)) ?
The solution to ((1+a^7))/((1+a)) is a^6-a^5+a^4-a^3+a^2-a+1