Been having some trouble with some equations lately. I would love to get a solution to some if not all of these.
ax(a − x) ^2 (= a ^3 x + ax^3 − 2a ^2 x ^2 )
y(b + y b) ^2 − y^3 b^2 (= y b^2 + 2y^2 b^2 )
y/4p + 3y/p (= 13y/4p)
((1/2)^3 + 1) ((1/2)^3 − 1) (= 2^−6 − 1)
(√2 + √8)^2 (= 18) †
6/11(1/ 3 +3/ 2) (= 1)
Expand the following:
a x (a-x)^2
(a-x) (a-x) = (a) (a) + (a) (-x) + (-x) (a) + (-x) (-x) = a^2-a x-a x+x^2 = a^2-2 a x+x^2:
a x a^2-2 a x+x^2
a x (a^2-2 a x+x^2) = a x a^2+a x (-2 a x)+a x x^2:
a x a^2-2 a x a x+a x x^2
a x a^2 = a^(1+2) x:
a^(1+2) x-2 a x a x+a x x^2
1+2 = 3:
a^3 x-2 a x a x+a x x^2
a x (-2) a x = -2 a^2 x^2:
a^3 x+-2 a^2 x^2+a x x^2
a x x^2 = a x^(1+2):
a^3 x-2 a^2 x^2+a x^(1+2)
1+2 = 3:
Answer: |a^3x - 2a^2x^2 + ax^3
Expand the following:
y (b+y b)^2-y^3 b^2
(b y+b) (b y+b) = (b y) (b y) + (b y) (b) + (b) (b y) + (b) (b) = b^2 y^2+b^2 y+b^2 y+b^2 = b^2+2 b^2 y+b^2 y^2:
y b^2+2 b^2 y+b^2 y^2-y^3 b^2
y (b^2+2 b^2 y+b^2 y^2) = y b^2+y×2 b^2 y+y b^2 y^2:
y b^2+2 y b^2 y+y b^2 y^2-y^3 b^2
y×2 b^2 y = 2 b^2 y^2:
y b^2+2 y^2 b^2+y b^2 y^2-y^3 b^2
y b^2 y^2 = y^(1+2) b^2:
y b^2+2 y^2 b^2+y^(1+2) b^2-y^3 b^2
1+2 = 3:
y b^2+2 y^2 b^2+y^3 b^2-y^3 b^2
Grouping like terms, y b^2+y^2×2 b^2+y^3 b^2-y^3 b^2 = 2 b^2 y^2+b^2 y+(b^2 y^3-b^2 y^3):
2 b^2 y^2+b^2 y+(b^2 y^3-b^2 y^3)
b^2 y^3-b^2 y^3 = 0:
Answer: | 2b^2y^2 + b^2y
Simplify the following:
y/(4 p)+(3 y)/p
Put each term in y/(4 p)+(3 y)/p over the common denominator 4 p: y/(4 p)+(3 y)/p = y/(4 p)+(12 y)/(4 p):
y/(4 p)+(12 y)/(4 p)
y/(4 p)+(12 y)/(4 p) = (y+12 y)/(4 p):
(y+12 y)/(4 p)
y+12 y = 13 y:
Answer: |13y / (4p)
((1/2)^3 + 1) * ((1/2)^3 − 1) = 2^−6 − 1 This is correct. Watch out for your brackets.
Simplify the following:
(sqrt(2)+sqrt(8))^2
sqrt(8) = sqrt(2^3) = 2 sqrt(2):
(sqrt(2)+2 sqrt(2))^2
sqrt(2)+2 sqrt(2) = 3 sqrt(2):
3 sqrt(2)^2
Multiply each exponent in 3 sqrt(2) by 2:
3^2×2^(2/2)
2/2 = 1:
3^2×2
3^2 = 9:
9×2
9×2 = 18:
Answer: |18
Simplify the following:
(6 (1/3+3/2))/(11)
(6 (1/3+3/2))/(11) = (6 (1/3+3/2))/(11):
(6 (1/3+3/2))/(11)
Put 1/3+3/2 over the common denominator 6. 1/3+3/2 = 2/6+(3×3)/6:
(6 2/6+(3×3)/6)/(11)
3×3 = 9:
(6 (2/6+9/6))/(11)
2/6+9/6 = (2+9)/6:
(6 (2+9)/6)/(11)
2+9 = 11:
(6×11/6)/11
6×11/6 = (6×11)/6:
(6×11)/6/11
((6×11)/6)/11 = (6×11)/(6×11):
(6×11)/(6×11)
(6×11)/(6×11) = 1:
Answer: | 1
