Let a,b and c are rtional numbers which satisfy a-7b+8c=4 an

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Let a,b and c are rtional numbers which satisfy a-7b+8c=4 and 8a+4b-c=7.Then a^2-b^2+c^2=?

有
a-7b+8c=4 (1)
8a+4b-c=7 (2)
(1)+(2)*8 可得: 65a+25b=60
化简得13a+5b=12 -->a=(12-5b)/13
(1)*8-(2)得-60b+65c=-25
化简得12b-13c=5 --->c=(12b+5)/13
将a,c,代入a²-b²+c²=((12-5b)/13)²+((12b+5)/13)²-b²
=((12-5b)²+(12b+5)²)/13²-b²
=(12²+(5b)²+(12b)²+5²)/13²-b²
=(12²+5²+(12²+5²)b²)/13²-b²
=1+b²-b²
=1

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