What is expand (1-t,3(1-t)(4+t),-(1-t+2-t))(-1,1,-1) ?

The solution to expand (1-t,3(1-t)(4+t),-(1-t+2-t))(-1,1,-1) is

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expand (1-t,3(1-t)(4+t),-(1-t+2-t))(-1,1,-1)

expand

(1 - t, 3 (1 - t) (4 + t), - (1 - t + 2 - t)) (- 1, 1, - 1)

- 1 \cdot 1, 1, - 1 \cdot 1 + t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16 + 1, 1, 1 \cdot 4 t^{2} - 1, 1, 1 \cdot 12 t + 1, 1, 1 \cdot 9

Solution steps

(1 - t, 3 (1 - t) (4 + t), - (1 - t + 2 - t)) (- 1, 1, - 1)

= (- 1 \cdot 1,, - 1) (1 - 3,, t (1 - t) (4 + t) - (1 - t + 2 - t))

Expand

(1 - t, 3 (1 - t) (4 + t), - (1 - t + 2 - t)) (- 1, 1, - 1) : - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3 (1 - t) (4 + t), 1, 1, t, 3 (1 - t) (4 + t), 1 + 1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t)

= - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3 (1 - t) (4 + t), 1, 1, t, 3 (1 - t) (4 + t), 1 + 1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t)

Expand

t, 3 (1 - t) (4 + t), 1, 1, t, 3 (1 - t) (4 + t), 1 : t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16

= - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16 + 1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t)

Expand

1 \cdot (1 - t + 2 - t), 1, 1 \cdot (1 - t + 2 - t) : 1, 1, 1 \cdot 4 t^{2} - 1, 1, 1 \cdot 12 t + 1, 1, 1 \cdot 9

= - 1 \cdot 1, 1, - 1 \cdot 1 + t, 3, 1, 1, t, 3, 1 \cdot t^{4} + t, 3, 1, 1, t, 3, 1 \cdot 6 t^{3} + t, 3, 1, 1, t, 3, 1 \cdot t^{2} - t, 3, 1, 1, t, 3, 1 \cdot 24 t + t, 3, 1, 1, t, 3, 1 \cdot 16 + 1, 1, 1 \cdot 4 t^{2} - 1, 1, 1 \cdot 12 t + 1, 1, 1 \cdot 9

\frac{2 x + 1}{U ^{6} ( x - 3 ) ^{6} x}

\frac{( ( x + 3 ) ^{2} )}{( x + 1 ) ^{3} ( x + 2 ) ^{2}}

(f (x) = 2, x + 5) \cdot (g (x) = x + 5)

25 (6 - 2)

- 3 (3 x - 5 = 7 - 4 x)

What is expand (1-t,3(1-t)(4+t),-(1-t+2-t))(-1,1,-1) ?
The solution to expand (1-t,3(1-t)(4+t),-(1-t+2-t))(-1,1,-1) is