if x^2+x+1=0 find velue of (ax^2+bx+c)^2+(bx^2+cx+a)^2+(cx^2+bx+a)^2=?
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Answer:
Answer:
=a2x4+b2x4+c2x4+2abx3+4bcx3+2abx2+4acx2+2b2x2+c2x2+2abx+2acx+2bcx+2a2+c2
Step-by-step explanation:
Let's simplify step-by-step.
(ax2+bx+c)2+(bx2+cx+a)2+(cx2+bx+a)2
Distribute:
=a2x4+2abx3+2acx2+b2x2+2bcx+c2+b2x4+2bcx3+2abx2+c2x2+2acx+a2+c2x4+2bcx3+2acx2+b2x2+2abx+a2
Combine Like Terms:
=a2x4+2abx3+2acx2+b2x2+2bcx+c2+b2x4+2bcx3+2abx2+c2x2+2acx+a2+c2x4+2bcx3+2acx2+b2x2+2abx+a2
=(a2x4)+(b2x4)+(c2x4)+(2abx3)+(2bcx3+2bcx3)+(2abx2)+(2acx2+2acx2)+(b2x2+b2x2)+(c2x2)+(2abx)+(2acx)+(2bcx)+(a2+a2)+(c2)
=a2x4+b2x4+c2x4+2abx3+4bcx3+2abx2+4acx2+2b2x2+c2x2+2abx+2acx+2bcx+2a2+c2
Answer:
=a2x4+b2x4+c2x4+2abx3+4bcx3+2abx2+4acx2+2b2x2+c2x2+2abx+2acx+2bcx+2a2+c2
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