What is ((3)^4-81)/((3)^2+8(3)+15) ?

The solution to ((3)^4-81)/((3)^2+8(3)+15) is 0

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((3)^4-81)/((3)^2+8(3)+15)

\frac{( 3 ) ^{4} - 81}{( 3 ) ^{2} + 8 ( 3 ) + 15}

0

Solution steps

\frac{( 3 ) ^{4} - 81}{( 3 ) ^{2} + 8 ( 3 ) + 15}

Solve

(3)^{4} - 81 : 0

Follow the PEMDAS order of operations

Calculate exponents

(3)^{4} : 81

(3)^{4}

(3)^{4} = 81

= 81

= 81 - 81

Add and subtract (left to right)

81 - 81 : 0

81 - 81

81 - 81 = 0

= 0

= 0

Solve

(3)^{2} + 8 (3) + 15 : 48

Follow the PEMDAS order of operations

Calculate exponents

(3)^{2} : 9

(3)^{2}

(3)^{2} = 9

= 9

= 9 + 8 (3) + 15

Multiply and divide (left to right)

8 (3) : 24

8 (3)

Apply rule:

(a) = a

(3) = 3

= 8 \cdot 3

Multiply the numbers:

8 \cdot 3 = 24

= 24

= 9 + 24 + 15

Add and subtract (left to right)

9 + 24 + 15 : 48

9 + 24 + 15

9 + 24 = 33

= 33 + 15

33 + 15 = 48

= 48

= 48

= \frac{0}{48}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

[(3 - 5 (2 - 4) + 10 \div (- 3 - 2))]

10 + 2 (9 - 5) - \frac{16}{8}

90 - (25 + (5 \times 2 - 1) + 3)

10 - (10 - (- 10 - (- 10) + 10) - 10 + 10 + 100)

(3 - 5) \times (8 - 1)

What is ((3)^4-81)/((3)^2+8(3)+15) ?
The solution to ((3)^4-81)/((3)^2+8(3)+15) is 0