Question

we have to subtract .

Plz help me...............​

Answer 1

$\huge\boxed{\fcolorbox{cyan}{grey}{Solution:-}}$.

➡a/c to questions.

↪let , x whose number to add 2a³+5a-a2-6 get , a²-a-a³+1.

↪so,

(x)+(2a³+5a-a²-6)=(a²-a-a³+1).

=>X = (a²-a-a³+1-2a³+5a-a²-6).

=>X= -3a³+4a-5.

➡Thus:-

♠ (-3a³+4a-5) to add (2a³+5a-a²-6) to get (a²-a-a³+1).

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Answer 2

Answer:

Let the number to be added be X

So A/Q,

$\red{x +( 2a ^{3} + 5a -a ^{2} - 6 )= (a^{2} - a - a ^{3} + 1)}$

So by finding the value of X we can get the answer that the the number which is to be added with 2a³+5a-a²-6 to get a²-a-a³+1

So let us solve it so as to get the required number

$= > x + 2a ^{3} + 5a - a ^{2} - 6 = a ^{2} - a - a ^{3} + 1 \\ \\ = > x =a^{2} - a - a ^{3} + 1 - 2a ^{3} - 5a + a^{2} + 6 \\ \\ = > x = - 3a ^{3} + 2a^{2} - 6a + 7$

So the required number is

$\boxed { \red{ 3a^{2} - 2a ^{2} + 6a - 7}}$