factorise x to the power 6 minus 7 x cube minus 8
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18
Hey...<>Sol:
x6 + 5x3 + 8
= (x)6+ 5x3 + 8 + x3 - x3 = x6 + 6x3 + 8 - x3
[By adding and subtracting x3 on both sides]
= (x2)3+ (2)3+ (-x)3 - 3 × (x2) × (2) × (-x)
According to the identity, x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= (x2 + 2 - x)[(x2)2 + (2)2 + (-x)2 - (x2) × (2) - (2) × (-x) - (x2) × (-x))
= (x2 - x + 2)(x4 + x2 + 4 - 2x2 + 2x + x3)
= (x2 - x + 2)(x4 + x3 - x2 + 2x + 4)
Thus, the facotrs of x6 + 5x3 + 8 are (x2 - x + 2)(x4 + x3 - x2 + 2x + 4).
Plzz mark as brainslist.
x6 + 5x3 + 8
= (x)6+ 5x3 + 8 + x3 - x3 = x6 + 6x3 + 8 - x3
[By adding and subtracting x3 on both sides]
= (x2)3+ (2)3+ (-x)3 - 3 × (x2) × (2) × (-x)
According to the identity, x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= (x2 + 2 - x)[(x2)2 + (2)2 + (-x)2 - (x2) × (2) - (2) × (-x) - (x2) × (-x))
= (x2 - x + 2)(x4 + x2 + 4 - 2x2 + 2x + x3)
= (x2 - x + 2)(x4 + x3 - x2 + 2x + 4)
Thus, the facotrs of x6 + 5x3 + 8 are (x2 - x + 2)(x4 + x3 - x2 + 2x + 4).
Plzz mark as brainslist.
Answered by
20
x^6-7x^3-8
⇒x^6-1x^3
⇒x^3
⇒x^6-1x^3
⇒x^3
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